A block of mass m slides along a floor while a force of magnitude F is applied to it at an angle θ as shown in figure. The coefficient of kinetic friction is μₖ. then, the block's acceleration 'a' is given by :
(g is acceleration due to gravity)

Solution

Physics sir · 4 steps

The force pulls the block up a little, so the floor presses less and friction drops with it.

1. Split the force. F points up and to the right at θ above the horizontal. Its horizontal part F cos θ drives the block forward. Its vertical part F sin θ lifts upward.
2. Normal reaction. There is no vertical motion, so N + F sin θ = mg. That gives N = mg - F sin θ, which is less than mg.
3. Horizontal equation. Kinetic friction μₖ N acts backward, against the motion. Newton's second law along the floor then reads:

   ma = F cos θ - μₖ(mg - F sin θ)

4. So. Dividing by m gives a = F/m cos θ - μₖ(g - F/m sin θ), which is option (b).

The idea to keep. Always find the normal reaction before friction. A force pulling upward at an angle reduces N below mg, and a force pushing downward increases it.

Answer from the NTA final answer key. Solution by PrepWiser.