Question

the figure below shows a box with mass m = 5.6 kg pulled up a ramp inclined at an angle of θ = 28° with a force of magnitude f = 44 n parallel to the ramp.
(a) if there is no friction between the box and the ramp, what is the magnitude of the boxs acceleration (in m/s²)?
(b) if instead the coefficient of kinetic friction between the box and the ramp is 0.11, what is the magnitude of the boxs acceleration (in m/s²)?

Explanation:

Step1: Analyze forces in no - friction case

According to Newton's second law $F_{net}=ma$. The net force along the ramp is $F - mg\sin\theta$. So $F - mg\sin\theta=ma_1$.

Step2: Solve for acceleration in no - friction case

We can re - arrange the formula $a_1=\frac{F - mg\sin\theta}{m}$. Given $m = 5.6$ kg, $F = 44$ N, $\theta=28^{\circ}$, and $g = 9.8$ m/s². First, calculate $mg\sin\theta=5.6\times9.8\times\sin(28^{\circ})\approx5.6\times9.8\times0.469 = 25.77$. Then $a_1=\frac{44 - 25.77}{5.6}=\frac{18.23}{5.6}\approx3.26$ m/s².

Step3: Analyze forces in friction case

The frictional force $f=\mu_kN$, where the normal force $N = mg\cos\theta$. The net force along the ramp is $F - mg\sin\theta-\mu_kmg\cos\theta=ma_2$.

Step4: Solve for acceleration in friction case

We can re - arrange the formula $a_2=\frac{F - mg\sin\theta-\mu_kmg\cos\theta}{m}$. Calculate $mg\cos\theta=5.6\times9.8\times\cos(28^{\circ})\approx5.6\times9.8\times0.883 = 48.37$. Then $f=\mu_kN=0.11\times48.37 = 5.32$. And $a_2=\frac{44 - 25.77-5.32}{5.6}=\frac{12.91}{5.6}\approx2.30$ m/s².

Answer:

(a) 3.26 m/s²
(b) 2.30 m/s²