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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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Consider a box of mass 10 kg being lifted upward by a person standing on the surface of the earth. At a given instant the box is being accelerated upwards with an acceleration of size 5 m/s$^2$. Find the force that the person exerts on the box at this instant.

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Answer: 148 N
Given, the mass of the object is m = 1 0 kg and that the object is being accelerated upward. We need to find the Force exerted.
Let $\vec{F_P}$ be the force of the person lifting the box, and $\vec{F_g}$ be the gravitational pull of the earth.
The forces are in the opposite direction along the y-axis:
$\Rightarrow \vec{F_P} = F_{PY} \hat{y} = -mg \hat{y}$ (negative sign as the earth's gravitational pull is downwards).
$\Rightarrow \vec{F} = \vec{F_P} + \vec{F_g} = ma_y$
$\Rightarrow F_{PY} = ma_y+mg = m (a_y+g) = 10 Kg (5 + 9.8) m/s^2 =148\;N$
answered Aug 19, 2014 by balaji.thirumalai

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