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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 7 / Problem 81P

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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Answer: A wooden block with mass 1.50 kg is placed against

Chapter 7, Problem 81P

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QUESTION:

Problem 81P

A wooden block with mass 1.50 kg is placed against a compressed spring at the bottom of an incline of slope 30.0o (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 6.00 m up the incline from A , the block is moving up the incline at 7.00 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and the incline is µk = 0.50. The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

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QUESTION:

Problem 81P

A wooden block with mass 1.50 kg is placed against a compressed spring at the bottom of an incline of slope 30.0o (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 6.00 m up the incline from A , the block is moving up the incline at 7.00 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and the incline is µk = 0.50. The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

ANSWER:

Solution 81P

Step 1:

Data given

Mass of the block $m\ =1.5\ kg\$

Velocity up the plane $v=\ 7\ m/s$

Angle of inclination $\theta{}=\ 3{0}^{0}$

Coefficient of kinetic friction ${\mu{}}_{k}=\ 0.50$

We need to find the force exerted on the spring by the block

It is given by

${F}_{\ }={\mu{}}_{k}\ mg\ \$

Substituting values we get

${F}_{\ }=0.50\times{}\ 1.5\ kg\times{}9.8\ \$

${F}_{\ }=\ 7.35\ N\$

Hence the force on the spring is $\ 7.35\ N$

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