A crate with a total mass of 18.0 kg is pulled by a person along the floor at a constant speed by a rope. The rope is inclined at 20 degree above the horizontal, and the crate moves 20.0 m on a horizontal surface. The coefficient of kinetic friction between the floor and the crate is 0.50.
And the questions are as follows:
a. Draw a free body diagram for the crate.
b. What is the tension in the rope?
c. How much work is done on the crate by the rope?
d. What is the energy lost due to friction?
ma=Fcosα –F(fr)
a=0
0=Fcosα – μN= Fcosα – μmg
(b) F = μmg/cosα =
(c) W= Fcosα•d =
(d) W(fr) = μmg•d =
Thanks!
20
a. To draw a free body diagram for the crate, we need to identify and represent all the forces acting on the crate. Let's label the diagram as follows:
^ Normal Force (N)
|
---------------
| |
| Crate |
|_____________|
|
Frictional force (f)
|
|
|
|
v
The forces acting on the crate are:
1. Normal force (N): This force acts perpendicular to the contact surface of the crate and the floor. It counterbalances the force exerted by the weight of the crate.
2. Weight of the crate (mg): This force acts downward and is equal to the mass of the crate (m) multiplied by the acceleration due to gravity (g ≈ 9.8 m/s^2).
3. Frictional force (f): This force opposes the motion of the crate and acts parallel to the contact surface. Its magnitude is given by the equation f = µN, where µ is the coefficient of kinetic friction and N is the normal force.
Note: Since the crate is being pulled at a constant speed, there is no net force in the horizontal direction. Therefore, there is no acceleration in the horizontal direction.
b. The tension in the rope can be found by considering the forces acting vertically. Since the crate is being pulled at a constant speed, the tension in the rope must be equal to the weight of the crate. Therefore, the tension in the rope is given by the equation T = mg.
c. The work done on the crate by the rope can be calculated using the equation W = Fd, where W is the work done, F is the applied force, and d is the distance over which the force is applied. In this case, the applied force is equal to the tension in the rope (T) and the distance is given as 20.0 m. Therefore, the work done on the crate by the rope is W = Td.
d. The energy lost due to friction can be calculated using the equation E = f × d, where E is the energy lost, f is the frictional force, and d is the distance over which the frictional force acts. In this case, the frictional force is given by f = µN (from the free body diagram), and the distance is given as 20.0 m. Therefore, the energy lost due to friction is E = µNd.
Now you can use the given values to calculate the answers to parts b, c, and d.