First slide

Kinetic friction

Question

A boy of mass 25kg slides down a rope hanging from the branch of a tree. If the force of friction against him is 50N, the boy’s acceleration is (g=10ms^–2)

Moderate

A

10ms^–2
B

12ms^–2
C

8ms^–2
D

2ms^–2

Solution

Net downward force acting on the boy = mg - f_k ⇒ (25 x 10 - 50)N = 200N
$\therefore$ Acceleration = $\frac {200}{25}=8ms^{-2}$

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