In Fig. 8-49, a block slides down an incline. As it moves from point A to point B, which are 3.0 m apart, force acts on the block, with magnitude 3.8 N and directed down the incline. The magnitude of the frictional force acting on the block is 5.3 N. If the kinetic energy of the block increases by 32 J between A and B, how much work is done on the block by the gravitational force as the block moves from A to B?
Gravity does work equal to the KE increase PLUS the work done aqainst friction.
The gravity work should equal the potential energy change
A 3.0-kg block moves up a 40° incline with constant speed under the action of a 26-N force acting up and parallel to the incline. What magnitude force must act up and parallel to the incline for the block to move down the incline at constant velocity?
To find the work done on the block by the gravitational force as it moves from point A to point B, we can use the formula for work:
Work = force * distance * cosine(theta)
where force is the magnitude of the force acting on the block, distance is the displacement of the block, and theta is the angle between the force vector and the displacement vector.
Given that the distance between point A and point B is 3.0 m, the force acting on the block is 3.8 N directed down the incline, and the frictional force acting on the block is 5.3 N, we can assume that the angle between the force vector and the displacement vector is the same as the angle of the incline.
Since the block is sliding down the incline, the gravitational force component parallel to the incline is responsible for the work done on the block. The magnitude of this force can be calculated using the formula:
Force_parallel = force_down_incline - frictional_force
= 3.8 N - 5.3 N
= -1.5 N (The negative sign indicates that the force is acting in the opposite direction to the displacement)
We can assume the angle of the incline is given by:
theta = tan^(-1)(frictional_force / force_down_incline)
= tan^(-1)(5.3 N / 3.8 N)
= 53.13 degrees
Now we can calculate the work done by the gravitational force using the formula:
Work = force_parallel * distance * cosine(theta)
= -1.5 N * 3.0 m * cos(53.13 degrees)
= -7.595 J
Therefore, the work done on the block by the gravitational force as it moves from point A to point B is approximately -7.595 J.
To find the work done on the block by the gravitational force as it moves from point A to point B, we can use the equation:
Work (W) = force (F) × distance (d) × cosine of the angle between the force and displacement vectors.
In this case, the force acting on the block due to gravity is its weight, which can be calculated using the formula:
Weight (W) = mass (m) × gravitational acceleration (g),
where the mass of the block (m) and the gravitational acceleration (g) are constants.
To solve this problem, we need to know the angle of the incline. Since the angle is not given, let's assume it's known.
Let's say the angle between the incline and the horizontal plane is θ.
The force acting on the block downhill (F1) can be calculated using the given magnitude of the force (3.8 N) and the angle θ:
F1 = 3.8 N × sin θ.
The frictional force (Ff) acting on the block is given as 5.3 N.
Therefore, the net force acting on the block downhill (Fnet) is:
Fnet = F1 - Ff.
Assuming the block moves without acceleration, this net force will be equal to the force required to overcome friction:
Fnet = F1 - Ff = 0.
Substituting the values, we have:
3.8 N × sin θ - 5.3 N = 0.
Solving this equation will give us the angle θ between the incline and the horizontal plane.
Once we have the angle θ, we can calculate the work done by the gravitational force using the given distance (3.0 m):
Work by gravity (W) = Weight × distance × cos θ.
Since the mass of the block is not given, we cannot directly calculate its weight. However, the increase in kinetic energy can be used to find the mass using the formula:
Change in kinetic energy (ΔKE) = work done on the block.
Therefore:
32 J = W.
Now we can substitute the values and solve for work.