A man sits with his back against the back of a chair, and he pushes a block of m

Question

A man sits with his back against the back of a chair, and he pushes a block of mass m = 4 kg straight forward on a table in front of him, with a constant force F = 28 N, moving the block a distance d = 0.7 m. The block starts from rest and slides on a low-friction surface. (Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.)

(a) How much work does the man do on the block?

(b) What is the final kinetic energy K of the block?

(c) What is the final speed v of the block?

(d) How much time t does this process take?

(e) Consider the system of the man plus the block: how much work does the chair do on the man?

(f) What is the internal energy change of the man? J Now suppose that the man is sitting on a train that is moving in a straight line with speed V = 18 m/s, and you are standing on the ground as the train goes by, moving to your right. From your perspective (that is, in your reference frame), answer the following questions.

(g) What is the initial speed vi of the block?


(h) What is the final speed vf of the block?


(i) What is the initial kinetic energy Ki of the block?


(j) What is the final kinetic energy Kf of the block?
J

(k) What is the change in kinetic energy K = Kf Ki?

(l) How far does the block move (x)?


(m) How much work does the man do on the block?

(n) How far does the chair move?

(o) Consider the system of the man plus the block: how much work does the chair do on the man?

(p) What is the internal energy change of the man?

Explanation / Answer

Here ,

mass of block , m = 4 Kg

F = 28 N

d = 0.7 m

a) work done = Force * displacement

work done = 28 * 0.7

work done = 19.6 J

b)

final kinetic energy = net work done

final kinetic energy = 19.6 J

c)

let the final speed is v

0.5 * 4 * v^2 = 19.6

v = 3.13 m/s

the final speed v of the block is 3.13 m/s

d)

let the time taken is t

acceleration , a = 28/4

a = 7 m/s^2

v = u + a * t

3.13 = 7 * t

t = 0.45 s

the time taken is 0.45 s

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