1
Question
A block of mass \(m = (2 ~\text{kg})\) is placed on a rough horizontal surface and is being acted upon by a time dependant force \(F (N) = 2t\) (where \(t\) is in second). The coefficient of static friction between the block and the horizontal surface is \(\mu_s = 0.20\). The frictional force \(f\) developed between the block and the surface versus force \(F (N)\) plot is as shown.
The velocity of the block at \(t=4~\text{s}\) will be
\( 7 / l l l l l l l l l l l l l l l l \)
A block of mass \(m = (2 ~\text{kg})\) is placed on a rough horizontal surface and is being acted upon by a time dependant force \(F (N) = 2t\) (where \(t\) is in second). The coefficient of static friction between the block and the horizontal surface is \(\mu_s = 0.20\). The frictional force \(f\) developed between the block and the surface versus force \(F (N)\) plot is as shown.
The velocity of the block at \(t=4~\text{s}\) will be
\( 7 / l l l l l l l l l l l l l l l l \)
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Solution
The correct option is D 3 m/s
Maximum friction force
fmax=μsmg
=(0.2)(2)(10)=4 N
At t=2 s,F(N)=4 N
Upto t=2 s, block is in rest. After that relative motion starts between block and surface and kinetic friction acts between surfaces. Now from figure, force due to friction is Fr=4−1=3
Acceleration after two second is a=Fm
a=2t−32
dvdt=2t−32
∫v0dv=∫42(2t−32)dt
v=3 m/s
Maximum friction force
fmax=μsmg
=(0.2)(2)(10)=4 N
At t=2 s,F(N)=4 N
Upto t=2 s, block is in rest. After that relative motion starts between block and surface and kinetic friction acts between surfaces. Now from figure, force due to friction is Fr=4−1=3
Acceleration after two second is a=Fm
a=2t−32
dvdt=2t−32
∫v0dv=∫42(2t−32)dt
v=3 m/s
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