Question

A trolley is being pulled up an inclined plane by a man sitting on it (as shown in the figure). He applies a force of $250\mathrm{N}$. If the combined mass of the man and trolley is $100\mathrm{kg}$, the acceleration of the trolley will be $[\sin 15°=0.26]$.

• A. $2.4\mathrm{m}/{\mathrm{s}}^2$
• B. $9.4\mathrm{m}/{\mathrm{s}}^2$
• C. $6.9\mathrm{m}/{\mathrm{s}}^2$
• D. $4.9\mathrm{m}/{\mathrm{s}}^2$

Answer 1

The correct option is D

$4.9\mathrm{m}/{\mathrm{s}}^2$

Step 1: Given data

Force applied on the trolley, $F=250N$

The combined mass of trolley and man, $M=100kg$

The inclination of the plane, $\theta =15°$

Step 2: Formula used

$F_{net}=ma$

Where $F_{net}$ is net force, $m$ is mass, and $a$ is acceleration.

$W=mg\sin \theta$

Where $W$ is Weight, $m$ is mass, $g$ is the acceleration due to gravity, and $\theta$ the angle of inclination

Step 3: Calculation

The net force applied by tension force on the trolley system

$$\begin{gathered} F_{net}=3F \\ {F}_{net}=3\times 250N \\ {F}_{net}=750N \end{gathered}$$

Now total weight along the inclined plane downwards

$$\begin{gathered} W=mg\sin \theta \\ W=100\times 9.8\times \sin 15° \\ W=980\times 0.26 \\ W=253.6N \end{gathered}$$

Now the net upward force along the inclined,

$$\begin{gathered} F_{net}=3F-W \\ {F}_{net}=750N-253.6N \\ {F}_{net}=496.35N \end{gathered}$$

By Newton's second law,

$$\begin{gathered} F_{net}=ma \\ 496.35N=100·a \\ a=\frac{496.35}{100} \\ a=4.96m/s^2 \end{gathered}$$

Hence, the acceleration of the trolley will be $4.9\mathrm{m}/{\mathrm{s}}^2$.

So, option (D) is the correct option.