What is the force exerted by $5kg$ block on the lift as shown in fig? ($g=10m{{s}^{-2}}$)
(A). $100N$
(B). $115N$
(C). $105N$
(D). $135N$
Answer
632.7k+ views
Hint: The system contains a lift, two blocks of masses 5kg and 2kg. An external force pulls the lift against gravity with acceleration. The main forces acting on the objects of the system are force of gravity and normal reactions. Using Newton’s second law, we can make equations and solve them to get the required value of normal reaction.
Formula used:
$N-mg=ma$
$N'-N-m'g=m'a$
Complete step-by-step solution:
The lift is moving against the force of gravity, this means that force is being done on it against gravity. External force is being applied on the system which accelerates it with acceleration $5m{{s}^{-2}}$.
The normal reaction is a force which acts between two surfaces in order to prevent them from passing through each other.
Newton's second law states that a force is required to change the state of rest or motion of a body. Therefore, applying Newton’s second law, we will make equations using forces from the FBD diagram of each block.
The lift carrying two weights is moving with $5m{{s}^{-2}}$ upwards. Therefore, both blocks are also moving upwards at $5m{{s}^{-2}}$
From the above figure, the forces acting on the $2kg$ block are-
$\begin{align}
& N-mg=ma \\
& \Rightarrow N-2\times 10=2\times 5 \\
& \Rightarrow N=10+20 \\
& \therefore N=30N \\
\end{align}$
The normal reaction between both blocks is $30N$.
From the above figure, the forces acting on block of $5kg$ is-
$\begin{align}
& N'-N-m'g=m'a \\
& \Rightarrow N'-30-5\times 10=5\times 5 \\
& \Rightarrow N'=25+50+30 \\
& \therefore N'=105N \\
\end{align}$
Therefore, the normal reaction acting between the block and lift is $105N$. Hence, the correct option is (C).
Note:
The normal reaction is always perpendicular to the surfaces. It follows Newton’s third law; this means the normal reaction on both surfaces is equal and opposite in nature. The force of gravity acts at the centre of mass. Here energy is not conserved as the force of gravity does work on the system.
Formula used:
$N-mg=ma$
$N'-N-m'g=m'a$
Complete step-by-step solution:
The lift is moving against the force of gravity, this means that force is being done on it against gravity. External force is being applied on the system which accelerates it with acceleration $5m{{s}^{-2}}$.
The normal reaction is a force which acts between two surfaces in order to prevent them from passing through each other.
Newton's second law states that a force is required to change the state of rest or motion of a body. Therefore, applying Newton’s second law, we will make equations using forces from the FBD diagram of each block.
The lift carrying two weights is moving with $5m{{s}^{-2}}$ upwards. Therefore, both blocks are also moving upwards at $5m{{s}^{-2}}$
From the above figure, the forces acting on the $2kg$ block are-
$\begin{align}
& N-mg=ma \\
& \Rightarrow N-2\times 10=2\times 5 \\
& \Rightarrow N=10+20 \\
& \therefore N=30N \\
\end{align}$
The normal reaction between both blocks is $30N$.
From the above figure, the forces acting on block of $5kg$ is-
$\begin{align}
& N'-N-m'g=m'a \\
& \Rightarrow N'-30-5\times 10=5\times 5 \\
& \Rightarrow N'=25+50+30 \\
& \therefore N'=105N \\
\end{align}$
Therefore, the normal reaction acting between the block and lift is $105N$. Hence, the correct option is (C).
Note:
The normal reaction is always perpendicular to the surfaces. It follows Newton’s third law; this means the normal reaction on both surfaces is equal and opposite in nature. The force of gravity acts at the centre of mass. Here energy is not conserved as the force of gravity does work on the system.
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