Question

1. jonny pulls his sister jane (weight 28 lbs), who is sitting in a wagon, up an incline ramp (θ = 17°) with a steady speed. if the coefficient of kinetic friction is 0.18, the wagon has a mass of 14 kg and the length of the ramp is 2.4 m, find: a) the work done by the frictional force. b) the work done by the gravitational force.

Explanation:

Step1: Convert weight to mass in SI - units

First, convert Jane's weight from pounds to kilograms. 1 lb = 0.453592 kg. So, $m_{Jane}=28\times0.453592\ kg\approx12.7\ kg$. The total mass of Jane and the wagon is $m = 12.7\ kg+ 14\ kg=26.7\ kg$.

Step2: Analyze forces perpendicular to the ramp

The normal force $N$ on the wagon - Jane system on the ramp is given by $N = mg\cos\theta$, where $g = 9.8\ m/s^{2}$ and $\theta = 17^{\circ}$. So, $N=26.7\times9.8\times\cos(17^{\circ})\ N$. $\cos(17^{\circ})\approx0.9563$, then $N = 26.7\times9.8\times0.9563\ N\approx250.7\ N$.

Step3: Calculate the frictional force

The frictional force $f=\mu_{k}N$, where $\mu_{k}=0.18$. So, $f = 0.18\times250.7\ N\approx45.1\ N$.

Step4: Calculate the work - done by the frictional force

The work - done by a force $W = Fd\cos\varphi$, where $F$ is the force, $d$ is the displacement, and $\varphi$ is the angle between the force and the displacement. For the frictional force, $\varphi = 180^{\circ}$ (since the frictional force opposes the motion), and $d = 2.4\ m$. So, $W_f=f\times d\times\cos(180^{\circ})$. Substituting the values, $W_f=45.1\times2.4\times(- 1)\ J=-108.24\ J$.

Step5: Analyze forces parallel to the ramp for gravitational force component

The component of the gravitational force along the ramp is $F_g = mg\sin\theta$. So, $F_g=26.7\times9.8\times\sin(17^{\circ})\ N$. $\sin(17^{\circ})\approx0.2924$, then $F_g=26.7\times9.8\times0.2924\ N\approx76.7\ N$.

Step6: Calculate the work - done by the gravitational force

The work - done by the gravitational force $W_g=F_g\times d\times\cos(180^{\circ})$ (since the gravitational - force component along the ramp opposes the upward motion), $W_g = 76.7\times2.4\times(-1)\ J=-184.08\ J$.

Answer:

a) The work done by the frictional force is $- 108.24\ J$.
b) The work done by the gravitational force is $-184.08\ J$.