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Chapter 5: Problem 48

In a circus performance, a monkey is strapped to a sled and both are given an initial speed of \(4.0 \mathrm{~m} / \mathrm{s}\) up a \(20^{\circ}\) inclined track. The combined mass of monkey and sled is \(20 \mathrm{~kg}\), and the coefficient of kinetic friction between sled and incline is \(0.20\). How far up the incline do the monkey and sled move?

Short Answer

Expert verified
The distance up the incline the monkey and sled move can be found using these calculations.

Step by step solution

01

Calculate the force of friction

The force of friction \(\(F_{friction}\)\) can be calculated as: \[\(F_{friction} = 渭mg\cos(胃)\]\] where \(渭\) is the coefficient of friction, \(m\) is the mass of the sled and monkey, \(g\) is the acceleration due to gravity, and \(胃\) is the angle of the incline. By substituting \(渭=0.20\), \(m=20kg\), \(g=9.8m/s^2\) and using \(cos (20^\circ)\), we can calculate the force of friction.
02

Calculate the net force

The net force \(\(F_{net}\)\) acting on the monkey and sled can be determined by considering both the component of the weight down the incline and the force of friction. The component of the weight down the incline is given by \(mg\sin(胃)\). Therefore, the net force, which is the difference between the component of the weight down the slope and the force of friction, can be given by: \[\(F_{net} = mg\sin(胃) - F_{friction}\]\] Substitute \(m=20kg\), \(g=9.8m/s^2\), use \(sin (20^\circ)\), and the earlier calculated force of friction into this equation to calculate the net force.
03

Use the equations of motion to find the distance

Knowing that the monkey and sled eventually come to a stop up the incline, we can use the equations of motion to find the distance \(s\). For an object moving under constant acceleration, the equation of motion connecting initial speed (\(u\)), final speed (\(v\)), acceleration (\(a\)), and distance (\(s\)) is: \[\(v^2 = u^2 + 2as\)\]. Rearranging this equation for \(s\) gives: \[\(s = (v^2 - u^2) / 2a\)\]. The initial speed \(u\) is 4.0 m/s, final speed \(v\) is 0 (since the monkey and the sled come to a stop), and the acceleration \(a\) is \(F_{net} / m\). Substitute these values into the equation to find \(s\).

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Most popular questions from this chapter

QIC The masses of the javelin, discus, and shot are \(0.80 \mathrm{~kg}, 2.0 \mathrm{~kg}\), and \(7.2 \mathrm{~kg}\), respectively, and record throws in the corresponding track events are about \(98 \mathrm{~m}, 74 \mathrm{~m}\), and \(23 \mathrm{~m}\), respectively. Neglecting air resistance, (a) calculate the minimum initial kinetic energies that would produce these throws, and (b) estimate the average force exerted on each object during the throw, assuming the force acts over a distance of \(2.0 \mathrm{~m}\). (c) Do your results suggest that air resistance is an important factor?

QiC A shopper in a supermarket pushes a cart with a force of \(35 \mathrm{~N}\) directed at an angle of \(25^{\circ}\) below the horizontal. The force is just sufficient to overcome various frictional forces, so the cart moves at constant speed. (a) Find the work done by the shopper as she moves down a \(50.0-\mathrm{m}\) length aisle. (b) What is the net work done on the cart? Why? (c) The shopper goes down the next aisle, pushing horizontally and maintaining the same speed as before. If the work done by frictional forces doesn't change, would the shopper's applied force be larger, smaller, or the same? What about the work done on the cart by the shopper?

A light spring with spring constant \(1.20 \times 10^{3} \mathrm{~N} / \mathrm{m}\) hangs from an elevated support. From its lower end hangs a second light spring, which has spring constant \(1.80 \times 10^{3} \mathrm{~N} / \mathrm{m}\). A \(1.50-\mathrm{kg}\) object hangs at rest from the lower end of the second spring. (a) Find the total extension distance of the pair of springs. (b) Find the effective spring constant of the pair of springs as a system. We describe these springs as being in series. Hint: Consider the forces on each spring scparately.

A sledge loaded with bricks has a total mass of \(18.0 \mathrm{~kg}\) and is pulled at constant speed by a rope inclined at \(20.0^{\circ}\) above the horizontal. The sledge moves a distance of \(20.0 \mathrm{~m}\) on a horizontal surface. The coefficient of kinetic friction between the sledge and surface is \(0.500\). (a) What is the tension in the rope? (b) How much work is done by the rope on the sledge? (c) What is the mechanical energy lost due to friction?

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