A block with a mass of m1 = 3 kg is pushed to the right with a force of 377 N fo

Question

A block with a mass of m1 = 3 kg is pushed to the right with a force of 377 N for a distance of 5.2 m across a horizontal frictionless surface, after which the force is removed. The 3-kg block then collides with a second block with a mass m2 = 1.1 kg, which is initially at rest. The two blocks stick together after the collision.

1) Determine the change in the internal energy of the system of blocks during the collision.

2)Now assume that the 1.1-kg block is traveling at 2 m/s to the left before the collision. The two blocks will still stick together after the collision. Determine the change in the internal energy of the system of blocks.

Explanation / Answer

1) Let us first calculate the initial energy of the first block v1i,

a1= F1/m1 = 377/3= 125.7 m/s^2

v1i ^2 = vi^2 +2*a1*d1

v1i^2 = 0^2 + 2*125.7*5.2    => v1i = 36.16 m/s

Now use law of conservation of energy to calculate common velocity (vc) after collision when two masses stick ,

m1v1i+m2v2i = m1v1f+m2v2f

m1v1i+m2v2i = (m1+m2)*vc

3.0*36.16 + 1.1*0 = (3.0+1.1)*vc    => 26.46 m/s

Change in internal energy = change in KE = KEf – KEi = [1/2*(m1+m2)*vc^2] -[ ½*m1v1i^2 + ½*m2*v2i^2] = [1/2*(3.0+1.1)*26.16^2] -[ ½*3.0*26.16^2 + ½*1.1*0^2] = 376.4 J

2) Now v2i= - 2m/s

gain use law of conservation of energy to calculate common velocity (vc) after collision when two masses stick ,

m1v1i+m2v2i = m1v1f+m2v2f

m1v1i+m2v2i = (m1+m2)*vc

3.0*36.16 + 1.1*-2.0 = (3.0+1.1)*vc    => 25.92 m/s

Change in internal energy = change in KE = KEf – KEi = [1/2*(m1+m2)*vc^2] -[ ½*m1v1i^2 + ½*m2*v2i^2] = [1/2*(3.0+1.1)* 25.92 ^2] -[ ½*3.0*26.16^2 + ½*1.1*2.0^2] = 348.57 J

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