Question

for both questions

A 0.36 kg object travels from point A to point B. If the speed of the object at point A is 7.0 m/s and the kinetic energy at point B is 8.0 J, determine the following (a) the kinetic energy (in )) of the object at point A (b) the speed (in m/s) of the object at point B m/s 17. (-/4 Points] DETAILS SERCPWA11 5.WA.019. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER A warehouse worker is pushing a 90.0 kg crate with a horizontal force of 278 N at a speed of v = 0.875 m/s across the warehouse floor. He encounters a rough horizontal section of the floor that is 0.75 m long and where the coefficient of kinetic friction between the crate and floor is 0.360. (a) Determine the magnitude and direction of the net force acting on the crate while it is pushed over the rough section of the floor. magnitude direction (b) Determine the net work done on the crate while it is pushed over the rough section of the floor. J (c) Find the speed of the crate when it reaches the end of the rough surface. m/s

Answer 1

1) Given:

m = 0.36 kg

v1 = speed at A = 7 m/s

K2 = kinetic energy at B = 8 J

a) Kinetic energy at point A = [answer]

mu * 0.36 * 72 = 8.82J 2 2

b) let the speed at B is v2.

Then, kinetic energy at B is .

K) mus 2

Therefore, [answer]

2K2 2*8 V2 6.67m/s m V 0.36

17) Given:

m = 90 kg

F = force applied = 278 N

v = initial speed = 0.875 m/s towards right.

L = length of rough section = 0.75 m

H= 0.36

a) friction force =

f = mg = 0.36 * 90 * 9.81 = 317.84N

Now, the force applied is towards the right, and the friction force is towards the left. As the value of friction force is greater than the applied force, the direction of the net force is in the direction of the force having the larger magnitude.

The magnitude of net force = f - F = 317.84 - 278 = 39.84 N [answer]

and the direction of net force is in the direction of friction force, that is towards left. [answer]

b) Net work done on the crate while moving on the rough surface is [-ve sign is used since the direction of net force is opposite to the direction of displacement]

W Fnet

= −39.84 * 0.75 = −29.88 J [answer]

c) Let $v_{f}$ be the speed at the end of the rough surface.

Applying work energy principle between starting and end of the rough surface:

Net work done = final kinetic energy - initial kinetic energy

-29.88 = =" 5ml mu

-m mos u mu? -mv2 – 29.88 2

► mv = mv2 – 2 * 29.88

2 * 29.88 — →v} = 02 m

[answer]

u= 0.8752 2 * 29.88 90 0.319m/s