Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 6, Problem 133P

(a)

To determine

The speed of the blocks moving along the inclines.

(a)

Expert Solution
Check Mark

Answer to Problem 133P

The speed of the blocks is 1.7m/s.

Explanation of Solution

Let us consider the distance be d moved along the incline by mass m2. Both masses move the same distance and have the same speed, since they are connected by a rope.

Write the equation for law of conservation of energy of the system.

K1i+K2i+U1i+U2i=K1f+K2f+U1f+U2f (I)

Here, K1i is the initial kinetic energy of the block 1, K2i is the initial kinetic energy of the block 2, U1i is the initial potential energy due to gravity of the block 1, U2i is the initial potential energy due to gravity of the block 2, K1f is the final kinetic energy of the block 1, K2f is the final kinetic energy of the block 2, U1f is the final potential energy due to gravity of the block 1, U2f is the final potential energy due to gravity of the block 2,

Write the equation for final kinetic energy.

K1f=12m1v2 (II)

Here, m1 is the mass of the block 1 and v is the velocity of the block.

Write the equation for final kinetic energy.

K2f=12m2v2 (III)

Here, m2 is the mass of the block 2 and v is the velocity of the block.

Write the equation for initial potential energy.

U1i=m1gdsinθ (IV)

Here, g is the gravitational acceleration, θ is the angle makes with the block 2, and d is the distance of the blocks travel.

Write the equation for final potential energy.

U2f=m2gdsinϕ (V)

Here, ϕ is the angle makes with the block 1.

Conclusion:

Substitute equation (II), (III), (IV) and (V) in equation (I).

0+0+m1gdsinθ+0=12m1v2+12m2v2+0+m2gdsinϕm1gdsinθ=12m1v2+12m2v2+m2gdsinϕm1gdsinθm2gdsinϕ=12m1v2+12m2v2

Solve the above equation for v.

gd(m1sinθm2sinϕ)=12v2(m1+m2)v2=2gd(m1sinθm2sinϕ)(m1+m2)v=2gd(m1sinθm2sinϕ)(m1+m2)

Substitute 9.80m/s2 for g, 2.00m for d, 6.00kg for m1, 36.9° for θ, 45.0° for ϕ, and 4.00kg for m2 in above equation.

v=2(9.80m/s2)(2.00m)(6.00kgsin36.9°4.00kgsin45.0°)(6.00kg+4.00kg)=(39.2m2/s2)(3.60kg2.83kg)(10.0kg)=1.7m/s

Therefore, the speed of the blocks is 1.7m/s.

(b)

To determine

Find the speed of block along the incline at constant acceleration.

(b)

Expert Solution
Check Mark

Answer to Problem 133P

The speed of block along the incline at constant acceleration is 1.7m/s.

Explanation of Solution

Write the equation for Newton’s second law applied for mass m1.

F=m1am1gsinθT=m1a (VI)

Here, T is tension exerted on the block 1 and a is the acceleration of the block.

Write the equation for Newton’s second law applied for mass m2.

F=m2aTm2gsinϕ=m2a (VII)

Rewrite the equation (VI) for tension force.

T=m1gsinθm1a (VIII)

Rewrite the equation (VII) for tension force.

T=m2gsinϕ+m2a (IX)

Compare the equation (VIII) and (IX) to solve acceleration.

m1gsinθm1a=m2gsinϕ+m2am1gsinθm2gsinϕ=m2a+m1ag(m1sinθm2sinϕ)=a(m1+m2)a=g(m1sinθm2sinϕ)(m1+m2) (X)

Write the equation for motion of blocks travel along the incline.

v2=u2+2ad (XI)

Here, v is the velocity of the block and u is the initial velocity of the block.

Conclusion:

Substitute 9.80m/s2 for g, 2.00m for d, 6.00kg for m1, 36.9° for θ, 45.0° for ϕ, and 4.00kg for m2 in equation (X).

a=9.80m/s2(6.00kgsin36.9°4.00kgsin45.0°)(6.00kg+4.00kg)=(9.80m/s2)(3.60kg2.83kg)(10.0kg)=0.76m/s2

Susbtitute 0m/s for u, 0.76m/s2 for a, and 2.00m for d in equation (XI)

v2=(0m/s)2+2(0.76m/s2)(2.00m)v2=3.04m2/s2v=1.7m/s

Therefore, the speed of block along the incline at constant acceleration is 1.7m/s.

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