Three part problem cant figure out the second and third parts
(part 1 of 3)
A cart loaded with bricks has a total mass
of 12.9 kg and is pulled at constant speed by
a rope. The rope is inclined at 20.6 ◦ above
the horizontal and the cart moves 25.8 m on
a horizontal floor. The coefficient of kinetic
friction between ground and cart is 0.2 .
The acceleration of gravity is 9.8 m/s2 .
What is the normal force exerted on the
cart by the floor? Answer in units of N.
(part 2 of 3)
How much work is done on the cart by the
rope? Answer in units of kJ.
(part 3 of 3)
Note: The energy change due to friction is a
loss of energy.
What is the energy change Wf due to fric-
tion? Answer in units of kJ.
For part 3, the energy change due to friction work will equal the product if the friction force and the distance moved, 25.8 m.
Well the answer to part one was 117.5 and do you have any idea about how to do part 2
To solve the second and third parts of this problem, we need to understand the concept of work and energy.
(part 1 of 3):
To find the normal force exerted on the cart by the floor, we need to consider the forces acting on the cart. In this case, the only vertical force is the weight of the cart, which is equal to its mass (12.9 kg) multiplied by the acceleration due to gravity (9.8 m/s^2). Therefore, the normal force exerted by the floor on the cart must be equal in magnitude and opposite in direction to counterbalance the weight of the cart. The normal force can be calculated as the weight of the cart, which is (12.9 kg) * (9.8 m/s^2) = 126.42 N.
(part 2 of 3):
To calculate the work done on the cart by the rope, we need to consider the force applied by the rope and the displacement of the cart. The force applied by the rope can be decomposed into two components: one in the horizontal direction, parallel to the displacement of the cart, and the other in the vertical direction, perpendicular to the displacement and cancelled out by the normal force.
The horizontal component of the force is responsible for the work done on the cart. The work done is given by the formula W = F * d * cos(θ), where F is the force acting on the cart, d is the displacement of the cart, and θ is the angle between the force and the displacement. In this case, the force applied by the rope is equal to the tension in the rope, and the displacement of the cart is given as 25.8 m.
To calculate the tension in the rope, we need to consider the horizontal equilibrium of forces. The horizontal component of the weight of the cart is balanced by the horizontal component of the force applied by the rope. Therefore, the horizontal component of the tension in the rope is equal to the weight of the cart, which is (12.9 kg) * (9.8 m/s^2).
Now, knowing the magnitude of the horizontal tension in the rope, we can calculate the work done on the cart. The work done is given by W = (Force * Displacement * cos(θ)). In this case, the Force is the horizontal component of the tension in the rope, and the displacement is 25.8 m.
(part 3 of 3):
The energy change due to friction is a loss of energy. The work done by friction can be calculated using the formula Wf = Force of Friction * Displacement, where the Force of Friction is given by the coefficient of kinetic friction (0.2) multiplied by the normal force (126.42 N). The displacement is given as 25.8 m.
To convert the answers to kilojoules, simply divide the answer by 1000, since 1 kilojoule is equal to 1000 joules.
I hope this explanation helps you solve the second and third parts of the problem.