Question

A $4.0kg$box is on a frictionless $35°$slope and is connected via a massless string over a massless, frictionless pulley to a hanging $2.0kg$weight. The picture for this situation is similar to FIGURE P$7.39$.

a. What is the tension in the string if the $4.0kg$box is held in place, so that it cannot move?

b. If the box is then released, which way will it move on the slope?

c. What is the tension in the string once the box begins to move?

Short answer

(a) Tension in the spring if $4kg$box is held is $19.6N$

(b) The moving direction o the box is anti-clockwise which is mass $m_1$side.

(c) Tension in the spring, while the box is moving, is $20.56N$

Step-by-step solution

Given information (part a)

$m_1=4kg$

$m_2=2kg$

$\theta =35°$

Explanation (part a)

$$\begin{gathered} F_{net}=ma \\ {F}_{net}=m_{1}g\sin \left(\theta \right)-m_{2}g \\ m=m_1+m_2 \\ a=\frac{m_{1}g\sin \left(\theta \right)-m_{2}g}{m_1+m_2} \\ a=\frac{\left(4kg\right)\left(9.8m/s^2\right)\left(\sin \left(35°\right)\right)-\left(2kg\right)\left(9.8m/s^2\right)}{4kg+2kg} \\ a=0.48m/s^2 \end{gathered}$$

If the $4kg$box is held in place, then the tension will be because of only $2kg$box.

Tension T,

localid="1649649415107" $$\begin{gathered} T=m_{2}g \\ =\left(2kg\right)\left(9.8m/s^2\right) \\ =19.6N \end{gathered}$$

Given information (part b)

$m_1=4kg$

$m_2=2kg$

$\theta =35°$

Explanation (part b)

If the box is released direction will be in the anti-clock direction because a is coming positive and we assume the direction of an in anti clock wise direction.

Given information (part c)

$m_1=4kg$

$m_2=2kg$

$\theta =35°$

Explanation (part c)

Tension in the string while box is moving

$$\begin{gathered} F_{net}=ma \\ {T}_2-m_{2}g=m_{2}a \\ {T}_2=m_{2}g+m_{2}a \\ a=0.48m/s^2 \\ {m}_2=2kg \\ {T}_2=\left(2kg\right)\left(9.8m/s^2\right)+\left(2kg\right)\left(0.48m/s^2\right) \\ =20.56N \end{gathered}$$