Question

A crate of mass 9.4 kg is pulled up a rough incline with an initial speed of 1.52 m/s. The pulling force is 110 N parallel to the incline, which makes an angle of 20.8° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 4.92 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 110-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 4.92 m?

Thank you so much for your help! An explanation of each would be greatly appreciated!!

Answer 1

a)
Work done by gravtitaional force = -increase in gravitational potential energy
= -m*g*h

h =l*sin 20.8 = 4.92* sin 20.8 = 1.75 m

Work done by gravtitaional force = -m*g*h
= -9.4*9.8*1.75
= -160.9 J

b)
Frictional force = miu*m*g*cos thetha
=0.4*9.4*9.8*cos 20.8
= 34.45 N

Increase in energy owing to friction = - work done by friction
= - ( - f * d)
Since f and d are in opposite direction , work done by friction will be negative
Increase in energy owing to friction = (f * d)
= (34.45*4.92)
           = 169.5 J

c)
Work done by 110 N= (F * d)
= (110*4.92)
= 541.2 J

d)
Final kinetic energy = initial kinetic energy + work done by friction + work done by 100 N + work done by gravity

Final kinetic energy - initial kinetic energy = work done by friction + work done by 100 N + work done by gravity
change in kinetic energy= work done by friction + work done by 100 N + work done by gravity
= -169.5 + 541.2 - 160.9
= 210.8 J

e)
Final kinetic energy - initial kinetic energy = 210.8 J
0.5*m*(Vf^2 - Vi^2) = 210.8
0.5*9.4*(Vf^2 - 1.52^2) =210.8
Vf= 6.86 m/s
Answer:6.86 m/s