Neglecting air friction, what is the acceleration of a stuntman who catches a tree while sliding down a slope inclined at angle θ, given a tension T?

To understand why the acceleration of the stuntman can be expressed as -T/m + g sin θ, we consider the forces acting on him as he slides down the slope.

The components of the forces acting on the stuntman include gravity and tension. The gravity acting down the slope can be broken down into two components based on the angle θ of the slope: one component acts parallel to the slope as g sin θ, and the other acts perpendicular to the slope and can be represented as g cos θ.

The tension T works in the opposite direction of the motion because it acts up the slope, opposing the parallel component of gravitational force. When the stuntman is sliding down the slope, the net force acting on him can be expressed as:

Net Force = (Force due to gravity down the slope) - (Tension)

In terms of the mathematical representation, the net force down the slope actually becomes:

F_net = m * a = m * g sin θ - T

To find acceleration ( a ), we could rearrange this equation to give:

a = (m * g sin θ - T) / m = g sin θ - T/m

To account for the direction of the forces correctly and to match conventions used in