# 1. Two forces F~ 1 and F~ 2 are acting on a block of mass m=1.5 kg. The magnitude of force F~ 1 is 12N and it makes an angle of θ = 37◦ with the horizontal as shown in figure-1. The block is sliding at a constant velocity over a frictionless floor.(a) Find the value of the normal force on the block.(b) Find the magnitude of force F~2 that is acting on the block(c) Find the magnitude of force F~ 2 if the block accelerates with a magnitude of a = 2.5 m/s2 along the direction of F~ 2 .

Normal force=7.48 N

Explanation:

N+F~1 sinθ-mg=0

=>N=1.5*9.8-12 sin37◦

=>N=14.7-7.22=7.48 N

## Related Questions

At one instant, a 17.0-kg sled is moving over a horizontal surface of snow at 4.10 m/s. After 6.15 s has elapsed, the sled stops. Use a momentum approach to find the magnitude of the average friction force acting on the sled while it was moving.

force = 11.33

Explanation:

given data:

sled mass = 17.0 kg

inital velocity (U) = 4.10 m/s

elapsed time (T) 6.15 s

final velocity (V) = 0

final momentum P2 = 0

Initial momentum of sledge is

from newton second law of motion

Kgm/s^2

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A hovering mosquito is hit by a raindrop that is 50 times as massive and falling at 8.5 m/sm/s , a typical raindrop speed. How fast is the raindrop, with the attached mosquito, falling immediately afterward if the collision is perfectly inelastic

Explanation:

The system mosquito-raindrop is described by the Principle of Momentum Conservation:

The final speed is:

The hormone glucagon is released by number of different tissues in the body to stabilize blood glucose levels. Which of the following pathways is least ikely to be activated by glucagon in hepatocytes? cells in the pancreas when blood sugar levels are low. It acts on a gluconeogenesis O glycogenolysis O ? oxidation glycolysis

Glycolysis

Explanation:

In human body, glucose levels are regulated by hormones  insulin and glucagon, secreted from pancreas. Glucagon from alpha cells and insulin from beta cells of the pancreas. Glucagon are regulated along depending upon the blood sugar levels. During fasting when blood sugar levels are decreased the glucagon levels are increased. Glucagon increases hepatic glucose through glycogenelysis.

So, Glycolysis of glucagon is least likely to be activated by glucagon in hepatocytes

You're carrying a 3.6-m-long, 21 kg pole to a construction site when you decide to stop for a rest. You place one end of the pole on a fence post and hold the other end of the pole 35 cm from its tip. For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution. Part A Part complete How much force must you exert to keep the pole motionless in a horizontal position? Express your answer in newtons. F = 114 N Previous Answers

This Physics problem involves balancing the forces and torques acting on a 3.6-m-long pole. By applying the principles of equilibrium and calculations of torque, we find that 114 N of force is needed to keep the pole in a horizontal position.

### Explanation:

This is a physics problem related to the concepts of equilibrium and torque. From the details provided, we know that the pole has a mass of 21 kg and it's 3.6 meters long. The center of gravity (cg) of the pole, since it's uniform, is at the middle, which is at 1.8 m from either end of the pole. We are then told that you are holding the pole 35 centimeters (or 0.35 meters) from its tip.

To keep the pole horizontal in equilibrium, the downward force due to the weight of the pole at its center of mass (which is equal to the mass of the pole times gravity, or 21*9.8 = 205.8 N) needs to be balanced by the sum of the torques produced by the forces you are applying at the end you are holding and the force exerted by the fence post at the other end.

Let the force you apply be F1 and the force the fence post exerts be F2. We have F2 at 0.35 m from one end (the pivot point), and F1 at 3.6 - 0.35 = 3.25 m from the pivot. Given that the torque (t) equals to Force (F) times the distance from the pivot (d), and that the net torque should equal zero in equilibrium, we have:

0.35*F2 = 3.25*F1   (1)

Because the net force should also be zero in equilibrium, we have:

F1 + F2 = 205.8    (2)

Solving these two equations, we'll be able to calculate that the force you must exert to keep the pole motionless in a horizontal position, F1, is approximately 114 N.

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To balance the 3.6m-long, 21 kg pole and keep it horizontally motionless, a force of approximately 114N is required

### Explanation:

The subject question is a classic example of Torque problem specific to Physics, which involves the concepts of force, weight, and distance. To keep the pole motionless and horizontally balanced, the force you exert must counterbalance the torque due to the pole's weight. Assuming the pole is uniform, its center of gravity (cg) is at its midpoint, 1.8m from each end. The weight of the pole acts downward at this midpoint, providing a clockwise torque about the point of support, which is the fence post.

This torque is calculated as Torque = r * F = 1.8m (distance from fence post to cg) * Weight of pole = 1.8m * 21kg * 9.8m/s² (gravitational acceleration) = ~370 N.m. As the pole is motionless, the total torque about any point must be zero. Hence, the counter-clockwise torque provided by the force you exert is equal to the clockwise torque due to the weight of the pole. Using the distance from the point of your hold to the fence post (3.25m) we can calculate the force you need to exert: Force = Torque/distance = 370 N.m/3.25m = ~114N.

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A cart with mass 340 g moving on a frictionless linear air track at an initial speed of 1.2 m/s undergoes an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart continues in its original direction at 0.66 m/s. (a) What is the mass of the second cart? (b) What is its speed after impact?

A) m2 = 98.71g

B) v_f2 = 1.86 m/s

Explanation:

We are given;

Mass of cart; m1 = 340g

Initial speed; v_i1 = 1.2 m/s

Final speed; v_f1 = 0.66 m/s

A)Since the collision is elastic, we can simply apply the conservation of momentum to get;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2) - - - - - (eq1)

From conservation of kinetic energy, we have;

(1/2)m1•(v_i1)² = (1/2)m1•(v_f1)² + (1/2)m2•(v_f2)² - - - - eq(2)

Let's make v_f2 the subject in eq 2;

Thus,

v_f2 = √([m1•(v_i1)² - m1•(v_f1)²]/m2)

v_f2 = √([m1((v_i1)² - (v_f1)²)]/m2)

Let's put this for v_f2 in eq1 to obtain;

m2 = {m1((v_i1) - (v_f1))}/√([m1((v_i1)² - (v_f1)²)]/m2)

Let's square both sides to give;

(m2)² = {m1•m2((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

This gives;

m2 = {m1((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

Plugging in the relevant values to get;

m2 = {340((1.2) - (0.66))²}/([(1.2)² - (0.66)²]

m2 = 98.71g

B) from equation 1, we have;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2)

Making v_f2 the subject, we have;

v_f2 = m1[(v_i1) - (v_f1)]/m2

Plugging in the relevant values to get;

v_f2 = 340[(1.2) - (0.66)]/98.71

v_f2 = 1.86 m/s

To determine the mass of the second cart and its speed after impact, we can use the principle of conservation of momentum. The initial momentum of the first cart is equal to its final momentum plus the momentum of the second cart. After calculating the mass of the second cart, we can use the conservation of momentum again to find its speed by equating the final velocity of the combined carts to the initial velocity of the first cart.

### Explanation:

To determine the mass of the second cart, we can use the principle of conservation of momentum. The initial momentum of the first cart, with a mass of 340 g and an initial velocity of 1.2 m/s, is equal to its final momentum plus the momentum of the second cart. Using this equation, we can solve for the mass of the second cart.

After calculating the mass of the second cart, we can use the conservation of momentum again to find its speed after the impact. Since the two carts stick together after the collision, the final velocity of the combined carts is equal to the initial velocity of the first cart. Using this equation, we can solve for the speed of the second cart.

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