An object of mass 20 kg is released at point A, slides to the bottom of the 30°

Question

An object of mass 20 kg is released at point A, slides to the bottom of the 30° incline, then collides with a horizontal massless spring, compressing it a maximum distance of 0.50 m (see below). The spring constant is 600 N/m, the height of the incline is 2.0 m, and the horizontal surface is frictionless.

What is the speed of the object (in m/s) at the bottom of the incline?

2.74 m/s

(b)

What is the work of friction (in J) on the object while it is on the incline?

-317 J

(c)

The spring recoils and sends the object back toward the incline. What is the speed of the object (in m/s) when it reaches the base of the incline?

2.74 m/s

(d)

What vertical distance (in m) does it move back up the incline?

____ M

Find part 4?

Explanation / Answer

(A) Applying energy conservation for mass-spring system,

k d^2 /2 = m v^2 / 2

600 (0.50^2) = 20 v^2

v = 2.74 m/s

(B) Applying work energy theorem,

work by gravity +work done by friction = changein KE

(20 x 9.81 x 2) + W = 20 x 2.74^2 /2 - 0

W = -317.4 J

(C) v = 2..74 m/s

(d) W = - 317.4 = f ( 2 / sin30)

f = 79.36 N

Applying work-energy theorem,

  
(-20 x 9.8 x h ) - (79.36 h / sin30) = 0 - 20(2.74^2)/2

h = 0.2 m .......Ans

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