A block of mass 3 kg starts from rest and slides down a curved path in the shape of a quarter-circle of radius 2 m and reaches the bottom of path with a speed 1 m/s . If g is 10 m/s2, the amount of work done against friction is

Question

A block of mass 3 kg starts from rest and slides down a curved path in the shape of a quarter-circle of radius 2 m and reaches the bottom of path with a speed 1 m/s . If g is 10 m/s2, the amount of work done against friction is (A) 60 J (B) 36 J (C) 24 J (D) 12 J

Tardigrade Admin · Accepted Answer

Applying Work-Energy theorem, we can write

Δ W = Δ K

\Rightarrow W_{t} + W_{g} = K_{t} = K_{i}

Where.

W_{f} =

Work done by frictional force

W_{g} =

Work done by gravity

W_{f} =

Final kinetic energy

W_{i} =

Initial kinetic energy

\Rightarrow W_{f} + m g = \frac{1}{2} m v^{2} - 0

\Rightarrow W_{f} = \frac{1}{2} m v^{2} - m g R

= \frac{1}{2} \times 3 \times 16 - 3 \times 10 \times 2

= 24 - 60 = - 36 J

Q. A block of mass 3kg starts from rest and slides down a curved path in the shape of a quarter-circle of radius 2m and reaches the bottom of path with a speed 1m/s . If g is 10m/s2, the amount of work done against friction is

60 J

36 J

24 J

12 J

Applying Work-Energy theorem, we can write ΔW=ΔK ⇒Wt​+Wg​=Kt​=Ki​ Where. Wf​= Work done by frictional force Wg​= Work done by gravity Wf​= Final kinetic energy Wi​= Initial kinetic energy ⇒Wf​+mg=21​mv2−0 ⇒Wf​=21​mv2−mgR =21​×3×16−3×10×2 =24−60=−36J

Q. A block of mass $3 k g$ starts from rest and slides down a curved path in the shape of a quarter-circle of radius $2 m$ and reaches the bottom of path with a speed $1 m / s$ . If $g$ is $10 m / s^{2},$ the amount of work done against friction is