Physics, 10.03.2020 08:15 sumayyahjj

Child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a speed of 0.75 m/s. The frictional force acting on the sled is one‑fifth of the combined weight of the child and the sled. If she travels for a distance of 25.5 m and her speed at the bottom is 4.30 m/s, calculate the angle that the slope makes with the horizontal.

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Answer from: ramzieboy13

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:

$\Delta K + \Delta U = W_{nc} (1)$

ΔK, is the change in kinetic energy, as follows:

$\Delta K = \frac{1}{2}* m* (v_{f} ^{2} - v_{0} ^{2}) (2)$

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:

$\Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)$

Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

$W_{nc} = F_{f} * d * cos 180\º \\\\ = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)$

Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

$\frac{1}{2}* (v_{f} ^{2} - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d$

Replacing by the givens and the knowns, we can solve for sin θ, as follows: $\frac{1}{2}*( (4.30 m/s) ^{2} - (0.75 m/s)^{2}) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = \frac{58.6 (m/s)2}{250 (m/s)2} = 0.236$ ⇒ θ = sin⁻¹ (0.236) = 13.7º

Answer from: Quest

anchor = archer //

Answer from: Quest

Answertraveling around a circular trackexplanation"traveling around a circular track" is the best answer as you could be traveling around a circular track at a constant speed. your velocity would therefore have the same magnitude but will always be constantly direction. since acceleration is the rate of change of velocity, the rate of change will only involve the direction of velocity and not the magnitude.wrong choices: "an object at rest" does not accelerate at all"increasing speed while traveling around a curve" means that the acceleration will not be constant due to only change in direction; it will be constant due to both magnitude and direction"an object in free fall" does not necessarily change direction

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