COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Question
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Chapter 6, Problem 133P

(a)

To determine

The velocity of the blocks moving after they travel 2.0m along the inclines.

(a)

Expert Solution
Check Mark

Answer to Problem 133P

The velocity of the blocks moving after they travel 2.0m along the inclines is 1.7m/s_.

Explanation of Solution

Write the expression for the distance travelled by the first block.

    d1=dsinθ        (I)

Here, d1 is the distance travelled by the first block, d is the distance along the incline, θ is the angle making the mass m1 between the horizontal and the inclined plane.

Write the expression for the distance travelled by the second block.

    d2=dsinϕ        (II)

Here, d2 is the distance travelled by the second block, ϕ is the angle making the mass m2 between the horizontal and the inclined plane.

Since the mass m1 is greater than the mass m2, the blocks slides along the incline to the right.

Write the expression for the conservation law of energy.

    12m1v2+12m2v2=m1gd1m2gd212(m1+m2)v2=m1gd1m2gd2        (III)

Here, v is the velocity of the blocks, g is the acceleration due to gravity.

Use equation (I) and (II) in (III) to solve for v.

    v=2[m1gdsinθm2gdsinϕ]m1+m2        (IV)

Conclusion:

Substitute 6.00kg for m1, 4.00kg for m2, 9.8m/s2 for g, 36.9° for θ, 45° for ϕ, 2.00m for d in equation (IV) to find v.

    v=2[(6.00kg)(9.8m/s2)(2.00m)sin36.9°(4.00kg)(9.8m/s2)(2.00m)sin45°](6.00kg+4.00kg)=5.51kgm2s210kg=1.7m/s

Therefore, the velocity of the blocks moving after they travel 2.0m along the inclines is 1.7m/s_.

(b)

To determine

The acceleration and velocity of each of the blocks.

(b)

Expert Solution
Check Mark

Answer to Problem 133P

The acceleration of each of the blocks is 0.76m/s2_ and velocity of each of the blocks is 1.7m/s_.

Explanation of Solution

COLLEGE PHYSICS-CONNECT ACCESS, Chapter 6, Problem 133P

Write the expression for the Newton’s second law of motion for the mass m2.

    F(m2)=m2aTm2gsinϕ=m2a        (V)

Here, F(m2) is the sum of forces acting on the mass m2, a is the acceleration of each blocks, T is the tension of the rope.

Write the expression for the Newton’s second law of motion for the mass m1.

    F(m1)=m1am1gsinθT=m1a        (VI)

Here, F(m1) is the sum of forces acting on mass m1.

Use equation (VI) to solve for T.

    T=m2gsinθm1a        (VII)

Use equation (VII) in (V) to solve for a.

    m1gsinθm1am2gsinϕ=m2am1gsinθm2gsinϕ=m2a+m1aa=g(m1sinθm2sinϕ)m1+m2        (VIII)

Initially the blocks are kept stationary and the initial velocity is zero.

Write the equation of motion to solve for the v.

    v2u2=2asv=2as        (IX)

Here, v is the final velocity, u is the initial velocity, s is the distance.

Conclusion:

Substitute 9.8m/s2 for g, 6.00kg for m1, 36.9° for θ, 4.00kg for m2, 45° for ϕ in equation (IX) to find a.

    a=(9.8m/s2)[(6.00kg)sin36.9°(4.00kg)sin45°](6.00kg+4.00kg)=0.76m/s2

Substitute 0.76m/s2 for a, 2.00m for s in equation (IX) to find v.

    v=2(0.76m/s2)(2.00m)=1.7m/s

Therefore, the acceleration of each of the blocks is 0.76m/s2_ and velocity of each of the blocks is 1.7m/s_.

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Chapter 6 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

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