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Question: A block of mass \(0.18kg\) is attached to a spring of force-constant \(2N/m\). The coefficient of fr...

A block of mass 0.18kg0.18kg is attached to a spring of force-constant 2N/m2N/m. The coefficient of friction between the block and the floor is 0.10.1. Initially, the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06m0.06m and comes to rest for the first time. The initial velocity of the block m/sm/s is V=N10V=\dfrac{N}{10}. Then N is:

A. 4
B. 5
C. 6
D. 7

Explanation

Solution

An object undergoing motion always follows the law of conservation of energy. According to the law, the energy is transformed from one form to another; it cannot be destroyed or created. This law of energy conservation can be used to find the initial velocity of the object.
Formula used:
KE=PE+frictionallossKE=PE+frictional\,loss

Complete answer:
As per the given data,
Mass is 0.18kg0.18kg
Spring constant 2N/m2N/m
The friction coefficient (μ)\left( \mu \right) is 0.1
Distance slide by the block (xx) is 0.6m0.6m
Initial velocity (uu) is N10m/s\dfrac{N}{10}m/s

As per the question a mass is attached with a spring which has it’s another end fixed at a certain point. Initially, both the block and string are at their original position. When an impulse is given to the block it slides a distance of 0.06m0.06m. After this, the object comes to rest.
So by considering the law of conservation of energy, kinetic energy will be transferred into other forms of the energy. The kinetic energy which the object has will transform into the potential energy and some of the energy will be undergoing frictional loss due to the friction between the two surfaces.
Mathematically,
KE=PE+frictionallossKE=PE+frictional\,loss
12mv2=12kx2+μmgx v2=kx2m+2μmgx \begin{aligned} & \Rightarrow \dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{2}k{{x}^{2}}+\mu mgx \\\ & \Rightarrow {{v}^{2}}=\dfrac{k{{x}^{2}}}{m}+2\mu mgx \\\ \end{aligned}
v=kx2m+2μmgx\Rightarrow v=\sqrt{\dfrac{k{{x}^{2}}}{m}+2\mu mgx}

By putting the values as per the given data,

& v=\sqrt{\dfrac{2{{(0.6)}^{2}}}{0.18}+2(0.1\times 0.6\times 10)} \\\ & \Rightarrow v=0.4 \\\ & \Rightarrow v=\dfrac{4}{10} \\\ \end{aligned}$$ From the given data it is known that the initial velocity of the block is given as $\dfrac{N}{10}m/s$ So it can be concluded that, $$v=\dfrac{4}{10}=\dfrac{N}{10}$$ By comparing we can come with the statement that the value of ‘N’ is 4. $N=4$ **So, the correct answer is “Option A”.** **Note:** The frictional force is the force which acts in the opposite direction of the applied force. It opposes the motion of the object. If an object has to be moved we have to apply a force greater than the frictional force (static frictional force).