A block of mass m slides along the track with kinetic friction mµ. A man pulls the block through a rope which makes an angle θ with the horizontal as shown in the figure. The block moves with a constant speed V. Power delivered by the man is:

Explanation

To determine the power delivered by the man, we need to consider the forces acting on the block and the work being done. Given: Mass of the block = m Coefficient of kinetic friction = µ The angle between the rope and the horizontal = θ Speed of the block = V The body is moving at constant speed, so the work is done by the horizontal component of the force. The horizontal component is T cos θ Power = Force × speed = T cosθ ( V ) = T V cos θ The forces acting on the block are: Gravity (mg), directed vertically downward Normal force (N), perpendicular to the track Kinetic friction force (f_k), opposing the motion of the block along the track. Tension in the rope (T), directed along the rope and making an angle θ with the horizontal. Since the block is moving at a constant speed V, the net force acting on the block in the horizontal direction is zero. Considering the horizontal forces, we have: T * cos(θ) - f_k = 0 The friction force can be calculated as: f_k = µN The normal force can be determined from the vertical forces: N + T * sin(θ) - mg = 0 V is constant and therefore the work is done by the horizontal component of the force i.e. Tcosθ Power= force X distance/time= Tcosθ X V (speed)= TVcosθ