In the figure below, a block of 1.67 slides on a track with different levels, which has friction only at the highest point where
the kinetic coefficient of friction is uk = 0.35. If the block has an initial speed V0 = 7.5m/s and the highest point of the track is at ℎ = 2.1 above the initial position of the block, calculate the distance where the friction force for the block is.
1 answer:
You might be interested in
Answer:
ANSWER BELOW I
I
V
Remember that w=mg where w is weight in Newtons, m is mass in kilograms, and g is gravity in
m/s2
. For example, for Earth, 445 N = 45.4 × 9.8
m/s2
:Notice that the x-axis values will be gravity in
m/s2
, which is already given in the table, and the y-axis values will be the weight in Newtons. Remember to round your weights to a whole number, and to enter the points starting with the lowest gravity (moon, then Mars, then Venus, then Earth).
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