a trunk is dragged across a level floor using a massless rope that makes an angle of 30 degrees with the horizontal. when the tension in the rope is 200 N and the kinetic friction coefficient = 0.5, the trunk moves at a constant speed. what is the mass of the trunk?
pulling force=200cos30
lifting force=200sin30
normal force=mg-lifting force
Net force=ma
in the horizontal, then if a=0
pulling force-friction=0
200cos30-.5(mg-200sin30)=0
solve for m
To find the mass of the trunk, we can use the concept of equilibrium for the forces acting on the trunk. In this scenario, the trunk is moving at a constant speed, which means the net force acting on it is zero.
Let's break down the forces acting on the trunk:
1. Tension in the rope (T): The given tension in the rope is 200 N, and it makes an angle of 30 degrees with the horizontal.
2. Weight of the trunk (W): The weight of an object is given by the formula W = m * g, where m is the mass and g is the acceleration due to gravity. Since the trunk is on a level floor, the vertical component of the weight is balanced by the normal force from the floor, leaving only the horizontal component of the weight.
3. Kinetic friction force (fk): The kinetic friction force is given by the formula fk = μk * N, where μk is the coefficient of kinetic friction and N is the normal force. Since the trunk is moving at a constant speed, the kinetic friction force is equal in magnitude and opposite in direction to the tension in the rope (T).
Now, let's calculate the mass of the trunk:
Since the trunk is moving at a constant speed, the net force acting horizontally is zero:
T * cos(30°) - fk = 0
T * cos(30°) = μk * N
T * cos(30°) = μk * m * g
200 N * cos(30°) = 0.5 * m * 9.8 m/s^2
Solving for m, we get:
m = (200 N * cos(30°)) / (0.5 * 9.8 m/s^2)
m ≈ 13.64 kg
Therefore, the mass of the trunk is approximately 13.64 kg.