Question

A crate of mass 10.2 kg is pulled up a rough incline with an initial speed of 1.56 m/s. The pulling force is 92 N parallel to the incline, which makes an angle of 19.0° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.04 m.

(a) How much work is done by the gravitational force on the
crate?

J

(b) Determine the increase in internal energy of the crate–incline
system owing to friction.

J

(c) How much work is done by the 92-N force on the crate?

J

(d) What is the change in kinetic energy of the crate?

J

(e) What is the speed of the crate after being pulled 5.04 m?

m/s

Answer #1

acceleration of the object

a= F/m - u g cos x - g sin x

=92/10.2 - 9.8* (0.4* cos 19 + sin 19)

= 9.02 - 6.897

a= 2.122 m/s^2

==========

a) workdone by gravitational force

W= mg sin x * d= 10.2*9.8* 5.04 sin 19= - 164.02 J

=========

b) workdone to change its internal energy

W= 0.4*10.2*9.8 cos 19* 5.04= - 190.54 J

==========

c) workdone by applied force

W= 92*5.04= 463.68 J

=======

d) using 3rd equation of motion

V^2= 2* 2.122* 5.04

V= 4.625 m/s

========

Comment in case any doubt.. good luck

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