A 100 lb block of ice slides down an incline 5.0 ft long and 3.0 ft high. A worker pushes up on the ice parallel to the incline so that it slides down at constant speed. The coefficient of friction between the ice and incline is 0.10.
Find the force exerted by the worker.
(I've found the force of the ice slipping down and friction force, but I have no idea how to find the worker's force...)
Also, how would you find the work done by gravity on the block? Is that a trick question?
i do not know
A 100 lb block of ice slides down an incline 5.0 ft long and 3.0 ft high. A worker pushes up on the ice parallel to the incline so that it slides down at constant speed. The coefficient of friction between the ice and incline is 0.10.
Find the force exerted by the worker.
(I've found the force of the ice slipping down and friction force, but I have no idea how to find the worker's force...)
A 100 lb block of ice slides down an incline 5.0 ft long and 3.0 ft high. A worker pushes up on the ice parallel to the incline so that it slides down at constant speed. The coefficient of friction between the ice and incline is 0.10.
Find a.the force exerted by the worker.
b.the work done by the man on the block
c.the work by the gravity on the block
the work by the surface of on the inclined on the block
A 100 lb block of ice slides down an incline 5.0 ft long and 3.0 ft high. A worker pushes up on the ice parallel to the incline so that it slides down at constant speed. The coefficient of friction between the ice and incline is 0.10.
Find a.the force exerted by the worker.
b.the work done by the man on the block
c.the work by the gravity on the block
the work by the surface of on the inclined on the block
To find the force exerted by the worker, we need to consider the forces acting on the block of ice. Let's break it down step by step.
First, let's identify the forces working on the block of ice sliding down the incline. We have:
1. The force of gravity acting downwards.
2. The normal force, which is the force exerted by the incline perpendicular to it.
3. The force of friction between the ice and the incline, acting parallel to the incline.
Since the ice is sliding at a constant speed, we know that the net force acting on it is zero. This means that the force exerted by the worker pushing up parallel to the incline must balance out the forces of gravity, normal force, and friction.
Let's calculate the force of gravity first:
The force of gravity (Fg) can be calculated using the formula:
Fg = m * g
where m is the mass of the block of ice and g is the acceleration due to gravity (which is approximately 32.2 ft/s^2).
In this case, the mass of the block of ice is given as 100 lb. Converting this to slugs (1 slug = 32.2 lb•s^2/ft), we have:
mass (m) = 100 lb / 32.2 lb•s^2/ft ≈ 3.105 slugs
Using the formula for force of gravity, we get:
Fg = 3.105 slugs * 32.2 ft/s^2 ≈ 100 lb
Next, let's calculate the force of friction:
The force of friction (Ff) can be calculated using the formula:
Ff = μ * N
where μ is the coefficient of friction and N is the normal force.
The normal force (N) can be calculated using the following equation:
N = m * g * cos(θ)
where θ is the angle of inclination.
In this case, the angle of inclination is not given, but we can determine it using the given height and length of the incline.
θ = arctan(height/length) ≈ arctan(3.0 ft/5.0 ft)
Using the arctan function or a calculator, we find that θ ≈ 30.96°.
Now, let's calculate the normal force (N):
N = 3.105 slugs * 32.2 ft/s^2 * cos(30.96°) ≈ 84.570 lb
Finally, we can calculate the force of friction (Ff):
Ff = 0.10 * 84.570 lb ≈ 8.457 lb
Since the net force is zero, the force exerted by the worker must balance out the force of gravity and the force of friction:
force exerted by the worker = Fg + Ff
force exerted by the worker = 100 lb + 8.457 lb ≈ 108.457 lb
Therefore, the force exerted by the worker is approximately 108.457 lb.