Question

A body of mass 3 kg is kept on a rough horizontal surface. A horizontal force of 20 N acting on the body produces an acceleration of $4 ms^{-2}$ in the body. To double the acceleration of the body, the horizontal force applied is to be increased by

• A. 100%
• B. 40%
• C. 60%
• D. 30%

Hint:

Be careful! Doubling the acceleration does {not} imply doubling the applied force when friction is involved. It doubles the {net} force. Always solve for the friction first.

Answer 1

The Correct Option is C

Step 1: Calculate the Frictional Force:
From Newton's Second Law, the net force causes acceleration: $F_{net} = ma$. Let the kinetic friction force be $f$. Given: Mass $m = 3$ kg, Initial Force $F_1 = 20$ N, Initial Acceleration $a_1 = 4 \, ms^{-2}$. \[ F_1 - f = ma_1 \] \[ 20 - f = 3(4) \] \[ 20 - f = 12 \implies f = 8 \text{ N} \] The kinetic friction remains constant (8 N) as long as the body is moving.
Step 2: Calculate the New Force Required:
We need to double the acceleration, so $a_2 = 2 \times 4 = 8 \, ms^{-2}$. Let the new applied force be $F_2$. \[ F_2 - f = ma_2 \] \[ F_2 - 8 = 3(8) \] \[ F_2 - 8 = 24 \implies F_2 = 32 \text{ N} \]
Step 3: Calculate the Percentage Increase:
Increase in applied force = $F_2 - F_1 = 32 - 20 = 12$ N. Percentage Increase: \[ % \text{ Increase} = \frac{\text{Change in Force}}{\text{Original Force}} \times 100 \] \[ % \text{ Increase} = \frac{12}{20} \times 100 \] \[ % \text{ Increase} = 0.6 \times 100 = 60% \]
Step 4: Final Answer:
The force needs to be increased by 60%.