In the figure, a person wants to rise a block lying on the ground to a height h. In both the cases, if the time required is same, then in which case he has to exert more force? Assume pulleys and strings light.

Question

In the figure, a person wants to rise a block lying on the ground to a height h. In both the cases, if the time required is same, then in which case he has to exert more force? Assume pulleys and strings light. (A) (i) (B) (ii) (C) Same in both (D) Cannot be determined

Tardigrade Admin · Accepted Answer

Since, h=(1/2) a t2, a should be same in both cases, because h and t are same in both cases as given. In Fig. (a), F1-m g=m a ⇒ F1=m g+m a In Fig. (b), 2 F2-m g=m a ⇒ F2=(m g+m a/2) ∴ F1>F2

Q. In the figure, a person wants to rise a block lying on the ground to a height $h$ . In both the cases, if the time required is same, then in which case he has to exert more force? Assume pulleys and strings light.

(i)

(ii)

Same in both

Cannot be determined

Since, $h = \frac{1}{2} a t^{2}, a$ should be same in both cases, because $h$ and $t$ are same in both cases as given.
In Fig. (a), $F_{1} - m g = ma$
$\Rightarrow F_{1} = m g + ma$
In Fig. (b), $2 F_{2} - m g = ma$
$\Rightarrow F_{2} = \frac{m g + ma}{2}$
$∴ F_{1} > F_{2}$

Q. In the figure, a person wants to rise a block lying on the ground to a height h. In both the cases, if the time required is same, then in which case he has to exert more force? Assume pulleys and strings light.

(i)

(ii)

Same in both

Cannot be determined

Since, h=21​at2,a should be same in both cases, because h and t are same in both cases as given. In Fig. (a), F1​−mg=ma ⇒F1​=mg+ma In Fig. (b), 2F2​−mg=ma ⇒F2​=2mg+ma​ ∴F1​>F2​

Q. In the figure, a person wants to rise a block lying on the ground to a height $h$ . In both the cases, if the time required is same, then in which case he has to exert more force? Assume pulleys and strings light.

Since, $h = \frac{1}{2} a t^{2}, a$ should be same in both cases, because $h$ and $t$ are same in both cases as given.
In Fig. (a), $F_{1} - m g = ma$
$\Rightarrow F_{1} = m g + ma$
In Fig. (b), $2 F_{2} - m g = ma$
$\Rightarrow F_{2} = \frac{m g + ma}{2}$
$∴ F_{1} > F_{2}$