Question

Chad got a new unpaid internship working in a coal mine pushing minecarts. All his minecarts

are loaded to weigh 340 kg.

(a) He first pushes a cart from rest on a frictionless section of track with a constant force of 300 N

for a distance of 10 m. How much work does he do on the cart? In what form is this energy?

What is the final speed of the cart?

(b) Then, starting from rest, he pushes another cart with the same 300 N a distance of 10 m up

a frictionless 5 degree slope (pushing in the parallel-to-surface direction). How much work

does he do on the cart? How much of this energy is in what forms?

(c) Now he pushes a cart from rest with force 300 N a distance of 10 m on a flat surface with

μs = μk = 0.05. How much work does he do on the cart? How much of this energy is in what

forms?

Answer 1

Given that :

weight on the minecart, m = 340 kg

(a) He first pushes a cart from rest on a frictionless section of track.

work on the cart which is given as :

W = F d cos                                                                            { eq.1 }

where, F = force = 300 N

d = distance = 10 m

inserting the values in eq.1,

W = (300 N) (10 m) cos (0)⁰

W = 3000 N

It is in the form of "kinetic energy". As,   W = K.E

the final speed of the cart which will be given as :

using equation of motion 3,    v² = v₀² + 2as                                              { eq.2 }

where, v₀ = 0 m/s

inserting the values in eq.2,

v² = (0 m/s)² + 2 (9.8 m/s²) (10 m)

v = 196 m²/s²

v = 14 m/s

(b) Then, starting from rest, he pushes another cart with the same force.

work on the cart which is given as :

W = F d cos    

where, = frictionless angle = 5 degree

then W = (300 N) (10 m) cos (5)⁰

W = 2988.3 N

It is in the form of "kinetic energy". As,   W = K.E K.E = 2988.3 N

(c) Now, he pushes a cart from rest with force 300 N a distance of 10 m on a flat surface. then, work done on the cart is given as :

W = f_k d cos                                                       

W = (_k F_N) d cos                                                                                       { eq.3 }

where, _k = 0.05

F_N = normal force on minecart = mg

= angle = 180⁰

inserting these values in eq.3,

W = [(0.05) (340 kg) (9.8 m/s²)] (10 m) cos 180⁰

W = -1666 N

It is in the form of "kinetic energy". As,   W = K.E K.E = -1666 N