A 100 N force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction μ= 0.5. If the acceleration due to gravity ( g ) is taken as 10 ms - 2 , the acceleration of the block (in ms -2 ) is

Question

A 100 N force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction μ= 0.5. If the acceleration due to gravity ( g ) is taken as 10 ms - 2 , the acceleration of the block (in ms -2 ) is (A) 2.5 (B) 10 (C) 5 (D) 7.5

Tardigrade Admin · Accepted Answer

a =( F -μ R / m )=(100-0.5 ×(10 × 10)/10)=5 ms -2

Q. A $100 N$ force acts horizontally on a block of $10 k g$ placed on a horizontal rough surface of coefficient of friction $μ =$ $0.5$ . If the acceleration due to gravity $(g)$ is taken as $10 m s^{-}$ 2 , the acceleration of the block (in $m s^{- 2}$ ) is

2.5

10

5

7.5

$a = \frac{F - μ R}{m} = \frac{100 - 0.5 \times ( 10 \times 10 )}{10} = 5 m s^{- 2}$

Q. A 100N force acts horizontally on a block of 10kg placed on a horizontal rough surface of coefficient of friction μ= 0.5. If the acceleration due to gravity (g) is taken as 10ms− 2 , the acceleration of the block (in ms−2 ) is

2.5

10

5

7.5

a=mF−μR​=10100−0.5×(10×10)​=5ms−2

Q. A $100 N$ force acts horizontally on a block of $10 k g$ placed on a horizontal rough surface of coefficient of friction $μ =$ $0.5$ . If the acceleration due to gravity $(g)$ is taken as $10 m s^{-}$ 2 , the acceleration of the block (in $m s^{- 2}$ ) is

$a = \frac{F - μ R}{m} = \frac{100 - 0.5 \times ( 10 \times 10 )}{10} = 5 m s^{- 2}$