Question

In: Physics

Consider two separate blocks with mass M1 and M2 on a horizontal frictionless surface, initially at rest. Both blocks are subjected to the same force of F

 

Consider two separate blocks with mass M1 and M2 on a horizontal frictionless surface, initially at rest. Both blocks are subjected to the same force of F (applied horizontally) and they are pushed D meters on the surface. If M1<M2, which one of the following is wrong?

  • A. Kinetic energy of block M1 is greater than the kinetic energy of block M2.
  • B. Speed of block M1 is greater than the speed of block M2.
  • C. Acceleration of block M1 is greater than the acceleration of block M2.
  • D. The work done on block M1 is equal to the work done on block M2.

Solutions

Expert Solution

Work-done = Force applied*displacement

In given scenario F force is applied on both blocks adn they are pushed D meters on the surface, So

W1 = Work-done on M1 = F*D

W2 = Work-done on M2 = F*D

So, W1 = W2

Option D is right.

Since both blocks are subjected to same force, So Using Newton's 2nd law:

F = M1*a1 = M2*a2

So, M1/M2 = a2/a1

Since given that: M1 < M2

So, M1/M2 < 1

which gives

a2/a1 < 1

a2 < a1

that means acceleration of block M1 is greater than acceleration of block M2

Option C is right.

Using Work-energy theorem:

W = dKE

W = KEf - KEi

Since Initially both blocks were at rest, So KEi = 0, So

W = KEf

Since Work-done on both block is same as discussed above, So

KE1 = KE2

(1/2)*M1*V1^2 = (1/2)*M2*V2^2

M1/M2 = (V2/V1)^2

Given M1 < M2

M1/M2 < 1

(V2/V1)^2 < 1

V2^2 < V1^2

V2 < V1

So Speed of block M1 is greater than speed of block M2

So option B is right.

As discussed above

KE1 = KE2

So option A is wrong,

Now since we need to find out wrong statement, So option A is wrong statement

 


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