Question

starting from rest, a 4.90 - kg block slides 2.40 m down a rough 30.0° incline. the coefficient of kinetic friction between the block and the incline is $mu_{k}=0.436$. (a) determine the work done by the force of gravity. (b) determine the work done by the friction force between block and incline. (c) determine the work done by the normal force. (d) qualitatively, how would the answers change if a shorter ramp at a steeper angle were used to span the same vertical height?

Explanation:

Step1: Calculate the vertical height for work - done by gravity

The vertical height $h$ the block descends along the incline of length $d = 2.40$ m and angle $\theta=30.0^{\circ}$ is $h = d\sin\theta$. The work - done by gravity $W_g$ is given by $W_g=mgh$, where $m = 4.90$ kg and $g = 9.8$ m/s².
$h=d\sin\theta=2.40\times\sin30.0^{\circ}=2.40\times0.5 = 1.2$ m
$W_g=mgh=4.90\times9.8\times1.2$
$W_g = 57.624$ J

Step2: Calculate the normal force and friction force for work - done by friction

The normal force $N$ on the block on the incline is $N = mg\cos\theta$. The friction force $f_k=\mu_kN=\mu_kmg\cos\theta$. The work - done by the friction force $W_f$ is $W_f=-f_kd=-\mu_kmg\cos\theta\times d$.
$N = mg\cos\theta=4.90\times9.8\times\cos30.0^{\circ}=4.90\times9.8\times\frac{\sqrt{3}}{2}\approx41.6$ N
$f_k=\mu_kN = 0.436\times41.6\approx18.1$ N
$W_f=-f_kd=-18.1\times2.40=-43.44$ J

Step3: Analyze work - done by normal force

The normal force is perpendicular to the direction of motion of the block. The work - done by a force $W = F\cdot d\cdot\cos\alpha$, where $\alpha$ is the angle between the force and the displacement. Since $\alpha = 90^{\circ}$ for the normal force and displacement along the incline, $\cos90^{\circ}=0$. So $W_N = 0$ J.

Step4: Qualitative analysis for steeper ramp

If a shorter ramp at a steeper angle is used to span the same vertical height:

• The work done by gravity $W_g=mgh$ depends only on the vertical height $h$ and the mass $m$ and the gravitational acceleration $g$. Since the vertical height is the same, the work done by gravity remains the same.
• The normal force $N = mg\cos\theta$. As $\theta$ increases, $\cos\theta$ decreases, so the normal force $N$ decreases. The friction force $f_k=\mu_kN$, so $f_k$ decreases. The distance $d$ along the ramp decreases as the ramp is shorter. The work done by friction $W_f=-f_kd$ decreases (becomes less negative) because both $f_k$ and $d$ decrease.
• The normal force is still perpendicular to the displacement along the ramp, so the work done by the normal force is still 0 J.

Answer:

(a) 57.6 J
(b) - 43.4 J
(c) 0 J
(d) Work done by gravity remains the same. Work done by friction decreases (becomes less negative). Work done by normal force remains 0 J.