Question

34. A block of mass 10 kg is released on rough incline plane. Block start descending with acceleration
2 m/s2. Kinetic friction force acting on block is (take g = 10 m/s2)
30°
(1)
10 N
(2) 30 N
(4) 50/3 N
(3) 50 N
​

Answer 1

Answer:

see the image posted ..............

Answer 2

Kinetic friction works on the block has magnitude 30N.

Explanation:

• Here, we can observe the situation and make a free body diagram according to the situation. (refer the figure attached)

• We can understand that the friction always opposes to the motion. It works opposite to the direction of motion.

Now, we apply the newton's law of motion and write as,

• $Net \ force = mass \times acceleration$

$mgsin \theta - (kinetic \ friction) = mass \times acceleration$

$mgsin30^o - (kinetic \ friction) = mass \times acceleration$

• Put the values of variables, we get;

$10 \times 10 \times \dfrac{1}{2} - (kinetic \ friction) = 10 \times 2$

• kinetic friction = 50 - 20 = 30 N

Thus, kinetic friction works on the block has magnitude 30N.