A 55-kg box is placed against a uniform spring with a spring constant (k) of 247

Question

A 55-kg box is placed against a uniform

spring with a spring constant (k) of 2470 N/m.

The box is pushed to the right so it

compresses the spring 20-cm. When the box

is released, it is propelled to the left.

(a) What is the velocity of the box as it slides along the flat surface to the left after it loses contact with the spring?

(b) How far up the ramp will the box slide while coming to a stop? Assume the flat section has no friction and the ramp has a coefficient of kinetic friction of .18 (c) After the box stops, will it remain stopped or will it slide back down the ramp (give rationale for your answer)?

The angle is 27 degrees

Explanation / Answer

Here ,

a)

let the velocity of the block is v

Using conservation of energy

0.50 * 55 * v^2 = 0.50 * 2470 * 0.20^2

v = 1.34 m/s

the speed of box is 1.34 m/s

b)

let the distance is d

d * 9.8 * ( sin(27 degree) + 0.18 * cos(27 degree)) = 0.50 * 1.34^2

solving for d

d = 0.149 m

the distance along the incline is 0.149 m

part c)

tan(theta) = tan(18 degree) = 0.32

hence , the block will slide back

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