A monkey of mass $50 \,kg$ climbs on a rope which can withstand the tension $(T)$ of $350 \,N$ If monkey initially climbs down with an acceleration of $4\, m / s ^2$ and then climbs up with an acceleration of $5 \,m / s ^2$ Choose the correct option $\left( g =10\, m / s ^2\right)$

Question

Find external force F so that block can move on inclined plane with constant velocity.

Answer 1

To solve this problem, we need to analyze the forces acting on the monkey while it climbs up and down using the rope.

When the monkey climbs down with an acceleration of 4 \, \text{m/s}^2, the net force acting on the monkey is given by: F_{\text{net, down}} = m(g - a_{\text{down}})
Here, m = 50 \, \text{kg}, g = 10 \, \text{m/s}^2, and a_{\text{down}} = 4 \, \text{m/s}^2.
The tension in the rope while climbing down is calculated as: T_{\text{down}} = m(g - a_{\text{down}}) = 50 \times (10 - 4) = 50 \times 6 = 300 \, \text{N}.
The tension 300 \, \text{N} is less than the maximum tension the rope can withstand, which is 350 \, \text{N}.
Conclusion: The rope will not break while the monkey is climbing down.

When the monkey climbs up with an acceleration of 5 \, \text{m/s}^2, the net force acting on it is: F_{\text{net, up}} = m(g + a_{\text{up}})
Here, a_{\text{up}} = 5 \, \text{m/s}^2.
The tension in the rope while climbing up is calculated as: T_{\text{up}} = m(g + a_{\text{up}}) = 50 \times (10 + 5) = 50 \times 15 = 750 \, \text{N}.
The tension 750 \, \text{N} exceeds the maximum tension the rope can withstand, which is 350 \, \text{N}.
Conclusion: The rope will break while the monkey is climbing upward.

Based on the calculations above, the correct answer is: Rope will break while climbing upward.

Answer 2

To solve this problem, we need to analyze the forces acting on the monkey while it climbs up and down using the rope.

When the monkey climbs down with an acceleration of 4 \, \text{m/s}^2, the net force acting on the monkey is given by: F_{\text{net, down}} = m(g - a_{\text{down}})
Here, m = 50 \, \text{kg}, g = 10 \, \text{m/s}^2, and a_{\text{down}} = 4 \, \text{m/s}^2.
The tension in the rope while climbing down is calculated as: T_{\text{down}} = m(g - a_{\text{down}}) = 50 \times (10 - 4) = 50 \times 6 = 300 \, \text{N}.
The tension 300 \, \text{N} is less than the maximum tension the rope can withstand, which is 350 \, \text{N}.
Conclusion: The rope will not break while the monkey is climbing down.

When the monkey climbs up with an acceleration of 5 \, \text{m/s}^2, the net force acting on it is: F_{\text{net, up}} = m(g + a_{\text{up}})
Here, a_{\text{up}} = 5 \, \text{m/s}^2.
The tension in the rope while climbing up is calculated as: T_{\text{up}} = m(g + a_{\text{up}}) = 50 \times (10 + 5) = 50 \times 15 = 750 \, \text{N}.
The tension 750 \, \text{N} exceeds the maximum tension the rope can withstand, which is 350 \, \text{N}.
Conclusion: The rope will break while the monkey is climbing upward.

Based on the calculations above, the correct answer is: Rope will break while climbing upward.

The Correct Option is C