A 6.30kg block is pushed 9.50m up a smooth 35.0∘ inclined plane by a horizontal force of 77.0N . If the initial speed of the block is 3.15m/s up the plane. What is the initial kinetic energy of the block, the work done by the 77.0N force, the work done by gravity, the work done by the normal force, and the final kinetic energy of the block?
I will be happy to check your thinking on this. on the horzontal force, you need to break that into normal up the ramp and up the ramp components.
To solve this problem, we can break it down into several steps. Let's go through each step to find the answers.
Step 1: Find the initial kinetic energy of the block.
The initial kinetic energy can be calculated using the formula:
Kinetic energy (KE) = 0.5 * mass * velocity^2
Given:
Mass of the block (m) = 6.30 kg
Initial velocity (v) = 3.15 m/s
Substituting the values into the formula:
KE = 0.5 * 6.30 kg * (3.15 m/s)^2
KE = 0.5 * 6.30 kg * 9.9225 m^2/s^2
KE ≈ 31.17 J
Therefore, the initial kinetic energy of the block is approximately 31.17 Joules.
Step 2: Find the work done by the 77.0N force.
To find the work done by a force, we need to multiply the force by the displacement along the force's direction.
Given:
Force applied (F) = 77.0 N
Displacement (s) = 9.50 m (up the inclined plane)
Work done (W) = Force * Distance * cos(theta)
where theta is the angle between the force and displacement vectors.
In this case, since the displacement is along the direction of the force, the angle theta is 0 degrees. Hence, cos(0) = 1.
W = 77.0 N * 9.50 m * cos(0)
W = 77.0 N * 9.50 m * 1
W = 731.5 Joules
Therefore, the work done by the 77.0N force is 731.5 Joules.
Step 3: Find the work done by gravity.
To find the work done by gravity, we need to multiply the gravitational force acting on the block by the vertical displacement.
Given:
Mass of the block (m) = 6.30 kg
Acceleration due to gravity (g) = 9.8 m/s^2
Displacement (s) = 9.50 m (up the inclined plane)
The force of gravity acting on the block is given by:
Force of gravity = mass * acceleration due to gravity
Hence, the work done by gravity is:
W = Force of gravity * displacement * cos(theta)
where theta is the angle between the force of gravity and displacement vectors.
Since the force of gravity acts vertically down while the displacement is along the inclined plane, the angle theta is 180 degrees. Hence, cos(180) = -1.
W = - (mass * acceleration due to gravity) * displacement * cos(180)
W = - (6.30 kg * 9.8 m/s^2) * 9.50 m * -1
W ≈ 579.3 Joules
Therefore, the work done by gravity is approximately 579.3 Joules.
Step 4: Find the work done by the normal force.
The normal force is perpendicular to the displacement, so no work is done by the normal force. Hence, the work done by the normal force is zero.
Step 5: Find the final kinetic energy of the block.
To find the final kinetic energy, we need to determine the final velocity of the block.
Since the block is pushed up a smooth inclined plane, there is no friction to slow it down or speed it up. Therefore, the work done by the force of 77.0N results in a change in potential energy, not kinetic energy.
Hence, the final kinetic energy is the same as the initial kinetic energy of the block, which we calculated as approximately 31.17 Joules in Step 1.
Therefore, the final kinetic energy of the block is approximately 31.17 Joules.