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Question

A 0.5 kg block slides from the point A (see figure) on a horizontal track with an inital speed of 3 m/s towards a weightless horizontal spring of length 1 m and force constant 2 N/m. The part AB of the track is fictionless and the part BC has the coefficients of static and kinetic friction as 0.22 and 0.2 respectively. If the distances AB and BD are 2 m and 2.14 m respectively, then

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Solution

The correct options are

**A** frictional force on the block when it comes to rest is 0.2 N.

**B** total distance travelled by the block is 4.24 m.

The part AB of the track is frictionless. Hence no energy is lost along AB. The 0.5 kg block travels to spring and compresses the spring by an amount x. Energy is stored in the spring.

Initial kinetic energy of block = (Work done against kinetic friction) + (potential energy of spring)

12Mv2=μkMg(2.14+x)+12kx2

or 12×0.5×(3)2

=0.2×0.5×10(2.14+x)+12×2x2

x=0.1 m

The spring gets compressed by 0.1m

∴ Restoring force exerted by spring on block

=2×0.1=0.2 N

Maximum possible value of static frictional force

=μSMg=(0.22)(0.5)(10)=1.1 N

Since this value is greater than restoring force.

∴ The block stops after compressing the spring by x.

∴ Total distance travelled by block =AB+BD+x

=2+2.14+0.1=4.24 m

The part AB of the track is frictionless. Hence no energy is lost along AB. The 0.5 kg block travels to spring and compresses the spring by an amount x. Energy is stored in the spring.

Initial kinetic energy of block = (Work done against kinetic friction) + (potential energy of spring)

12Mv2=μkMg(2.14+x)+12kx2

or 12×0.5×(3)2

=0.2×0.5×10(2.14+x)+12×2x2

x=0.1 m

The spring gets compressed by 0.1m

∴ Restoring force exerted by spring on block

=2×0.1=0.2 N

Maximum possible value of static frictional force

=μSMg=(0.22)(0.5)(10)=1.1 N

Since this value is greater than restoring force.

∴ The block stops after compressing the spring by x.

∴ Total distance travelled by block =AB+BD+x

=2+2.14+0.1=4.24 m

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