A block with mass m = 17 kg rests on a frictionless table and is accelerated by

Question

A block with mass m = 17 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 4215 N/m after being compressed a distance x1 = 0.52 m from the spring’s unstretched length. The floor is frictionless except for a rough patch a distance d = 2.1 m long. For this rough path, the coefficient of friction is µk = 0.45.
1)How much work is done by the spring as it accelerates the block?
2)What is the speed of the block right after it leaves the spring?
3)How much work is done by friction as the block crosses the rough spot?
4)What is the speed of the block after it passes the rough spot?

Explanation / Answer

Elastic potential energy stored in spring = 0.5kx^2= 0.5( 4215)( 0.52)^2 = 569 . 868J

Kinetic energy imparted to mass = 569 . 868J

a) Work done by spring = 569 . 868J

b) 0.5mv^2 = 569 . 868J

v = 8.2 m/s apprx

c) Work done by frictiomn= force of friction x displacemnt =- 0.45( 17)(9.8) x 2.1= -157.437 J apprx

d) retardation experiencedc by block = g = 0.45(9.8) = 4.41 m/s^2

v^2 - u^2 = 2as

v^2 = 8.2^2 - 2( 4.41)(2.1)

v= sqroot( 67.24 - 18.522) = 7 m/s apprx ( after moving out of rough patch)

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