A 6.15 kg block slides 1.49 m down a frictionless incline that makes an angle of 60.0° with the horizontal. What is the total work done on the block
what is the change in energy? Answer: mgh
so what is h? h=1.49sin60
To find the total work done on the block, we need to consider the work done by the gravitational force and the work done by the normal force.
1. Work done by the gravitational force:
The gravitational force is given by the formula: F_gravity = m * g * sin(θ), where m is the mass of the block (6.15 kg), g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of the incline (60.0°). The sin(θ) term represents the component of the gravitational force acting parallel to the incline.
The work done by the gravitational force is given by the formula: W_gravity = F_gravity * d, where d is the distance the block slides along the incline (1.49 m). Since the gravitational force is acting downhill, the work done by the gravitational force is negative.
Calculating the work done by the gravitational force:
F_gravity = (6.15 kg) * (9.8 m/s^2) * sin(60.0°)
W_gravity = F_gravity * (-1.49 m)
2. Work done by the normal force:
The normal force is the perpendicular force exerted by the incline on the block. In this case, since the incline is frictionless, there is no work done by the normal force because there is no displacement in the direction of the normal force.
3. Total work done:
To find the total work done, we add the work done by the gravitational force to the work done by the normal force:
Total work done = W_gravity + W_normal
Since W_normal is zero, the total work done is equal to the work done by the gravitational force:
Total work done = W_gravity
Now you can substitute the values into the equations and calculate the total work done on the block.