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A 4.0kgbox is on a frictionless 35°slope and is connected via a massless string over a massless, frictionless pulley to a hanging 2.0kgweight. The picture for this situation is similar to FIGURE P7.39.

a. What is the tension in the string if the 4.0kgbox is held in place, so that it cannot move?

b. If the box is then released, which way will it move on the slope?

c. What is the tension in the string once the box begins to move?

Short Answer

Expert verified

(a) Tension in the spring if 4kgbox is held is 19.6N

(b) The moving direction o the box is anti-clockwise which is mass m1side.

(c) Tension in the spring, while the box is moving, is 20.56N

Step by step solution

01

Given information (part a)

m1=4kg

m2=2kg

θ=35°

02

Explanation (part a)

Fnet=maFnet=m1gsinθ-m2gm=m1+m2a=m1gsinθ-m2gm1+m2a=4kg9.8m/s2sin35°-2kg9.8m/s24kg+2kga=0.48m/s2

If the 4kgbox is held in place, then the tension will be because of only 2kgbox.

Tension T,

localid="1649649415107" T=m2g=2kg9.8m/s2=19.6N

03

Given information (part b)

m1=4kg

m2=2kg

θ=35°

04

Explanation (part b)

If the box is released direction will be in the anti-clock direction because a is coming positive and we assume the direction of an in anti clock wise direction.

05

Given information (part c)

m1=4kg

m2=2kg

θ=35°

06

Explanation (part c)

Tension in the string while box is moving

Fnet=maT2-m2g=m2aT2=m2g+m2aa=0.48m/s2m2=2kgT2=2kg9.8m/s2+2kg0.48m/s2=20.56N

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