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Question: A block of mass 10 kg is placed on rough horizontal surface whose coefficient of friction is 0.5. If...

A block of mass 10 kg is placed on rough horizontal surface whose coefficient of friction is 0.5. If a horizontal force of 100 N is applied on it, then acceleration of the block will be (Take g=10ms−2g = 10ms^{- 2}) :

A

10ms−210ms^{- 2}

B

5ms−25ms^{- 2}

C

15ms−215ms^{- 2}

D

0.5ms−20.5ms^{- 2}

Answer

5ms−25ms^{- 2}

Explanation

Solution

Here, m=10kg,g=10ms−2,μ=0.5m = 10kg,g = 10ms^{- 2},\mu = 0.5

F =100 N

Force of friction

f=μN=μmg=0.5×10kg×10ms−2=50Nf = \mu N = \mu mg = 0.5 \times 10kg \times 10ms^{- 2} = 50N

Force that produces acceleration

F′=F−f=100N−50N=50NF^{'} = F - f = 100N - 50N = 50N

a=F′m=50N10kg=5ms−2a = \frac{F'}{m} = \frac{50N}{10kg} = 5ms^{- 2}