A crate of mass 9.6 kg is pulled up a rough incline with an initial speed of 1.46 m/s. The pulling force is 102 N parallel to the incline, which makes an angle of 19.9° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.02 m. a. How much work is done by the gravitational force on thecrate?b. Determine the increase in internal energy of the crate-inclinesystem owing to friction. c. How much work is done by the 100N force on the crate?d. What is the change in kinetic energy of the crate?e. What is the speed of the crate after being pulled 5.00m?

Answer with Explanation:

We are given that

Mass of crate,m=9.6 kg

Speed,u=1.46 m/s

Force,F=102 N

Distance,d=5.02 m

a.Work done=

Where

b.Increase in internal energy of the crate,E=

c.Work done by 102 N=

d.Change in kinetic energy=-160.7+512.04-177.75=173.6 J

e.

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s