Two workers are sliding 350 kgkg crate across the floor. One worker pushes forward on the crate with a force of 390 NN while the
1 answer:
Answer:
Explanation:
First, we write the equations of motion for each axis. Since the crate is sliding with constant speed, its acceleration is zero. Then, we have:
Where T is the tension in the rope, F is the force exerted by the first worker, f_k is the frictional force, N is the normal force and mg is the weight of the crate.
Since and
, we can rewrite the first equation as:
Now, we solve for and calculate it:
This means that the crate's coefficient of kinetic friction on the floor is 0.18.
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Answer:
θ = 13.7º
Explanation:
- According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
- The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
- As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:
- ΔK, is the change in kinetic energy, as follows:
- ΔU, is the change in the gravitational potential energy.
- If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:
- Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:
- Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:
- Replacing by the givens and the knowns, we can solve for sin θ, as follows:
⇒ θ = sin⁻¹ (0.236) = 13.7º