Question

A 4 kg box is on a frictionless 35° slope and is connected via a massless string over a massless, frictionless pulley to a hanging 2 kg weight, (a) What is the tension in the string if the 4 kg 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?

Answer 1

Answer:

(a) 19.62 N

(b) Box moves down the slope

(c) 24.43 N

Explanation:

(a)  

2 Kg box  causes tension

$T=mgwhere m is mass, g is gravitational force taken as 9.81T=2*9.81 =19.62 N (b) Block mass of 4 Kg [tex]T'-mg sin \theta=0$ hence $T'=mg sin \theta$ where m is mass and g is gravitational force  

T'=4*9.81 sin 35= 22.5071 N  

Since T' is greater than $mg sin\theta$ , then the box moves down the slope  

(c)  

Acceleration a= $\frac {forward force-backward force}{Total mass}= \frac {mg sin \theta -mg}{m1 + m2}$  

$a= \frac {22.51-19.62}{2+4}=0.48$

When moving, the box will exert force T"= $mgsin \theta + ma$  

T"= 4*9.81 sin 35 +(4*0.48)= 24.43 N

Answer 2

Answer:

(a). The tension in the string if the 4 kg box held in place is 22.48 N.

(b). The 4 kg box moves on downward.

(c). The tension in the string once the box begins to move is 24.4 N.

Explanation:

Given that,

Mass of box m₁ =4 kg

Mass of second box m₂= 2 kg

Angle = 35°

(a). We need to calculate the tension in the string if the 4 kg box is held in place, so that it cannot move

Using formula of tension

$T=mg\sin\theta$

Put the value into the formula

$T=4\times9.8\times\sin35$

$T=22.48\ N$

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

We need to calculate the tension for block of mass 2 kg

$T'=mg$

put the value into the formula

$T'=2\times9.8$

$T'=19.6\ N$

Here, T > T'

So, the first block moves on downward.

(c). We need to calculate the acceleration

Using formula of acceleration

$a =\dfrac{forward\ force-bacward\ force}{total\ mass}$

$a=\dfrac{T-T'}{M}$

Put the value into the formula

$a=\dfrac{22.48-19.6}{6}$

$a=0.48\ m/s^2$

We need to calculate the tension in the string once the box begins to move

For mass 4 kg

Using balance equation

$T-T''=ma$

$T''=T+ma$

Put the value into the formula

$T''=22.48+4\times0.47$

$T''=24.4\ N$

Hence, (a). The tension in the string if the 4 kg box held in place is 22.48 N.

(b). The 4 kg box moves on downward.

(c). The tension in the string once the box begins to move is 24.4 N.