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 (applied horizontally) and they are pushed D meters on the surface. If M1<M2, which one of the following is wrong?
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