Question

# A box with 10 kg of mass slides down an inclined plane that is 1.7 m...

A box with 10 kg of mass slides down an inclined plane that is 1.7 m high and 3.5 m long. Due to friction the box reaches 3.0 m/s at the bottom of the inclined plane. Beyond the inclined plane lies a spring with 650 N/m constant. It is fixed at its right end. The level ground between the incline and the spring has no friction

The box compressed the spring, got pushed back towards the incline by the spring. How far along the inclined plane, from the bottom, will the box temporarily stop on the inclined plane?

here,

mass of box , m = 10 kg

height of incline , h1 = 1.7 m

s1 = 3.5 m

theta = arcsin(h1/s1) = 29.1 degree

v = 3 m/s

let the friction force acting be ff

using Work energy theorm

m * g * h1 - ff * s1 = 0.5 * m * v^2

10 * 9.81 * 1.7 - ff * 3.5 = 0.5 * 10 * 3^2

solving for ff

ff = 34.8 N

when the block moves up the incline

let the block traveles s2 m before temporarily stop

using Work energy theorm

- m * g * s2 * sin(theta) - ff * s2 = (0 - 0.5 * m * v^2)

10 * 9.81 * s2 * sin(29.1) + 34.8 * s2 = 0.5 * 10 * 3^2

solving for s2

s2 = 0.55 m

the block traveles 0.55 m before temporarily stop

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