Question

A 750 kg car slows to a stop from traveling southbound at 30 m/s over a distance of 500 m. What was the acceleration needed to stop the car? How large was the force required to stop the car?

Two cables are being used to pull a 100 kg sled on a frictionless surface. The first cable has a tension of 500 N in a direction 15^o south of east. The other cable has a tension of 800 N in a direction 30^o east of north. What is the total horizontal force on the sled?

A 50 kg crate is being lowered by a cable. (a) If it is held still, what is the tension in the cable? (b) If it is being lowered with an acceleration of 1.0 m/s², what is the tension in the cable? (c) If it is being raised with an velocity of 1.0 m/s , what is the tension in the cable?

Answer 1

A)

mass of car,$m=750kg$

initial velocity ,$u=30m/s$

final velocity ,$v=0m/s$

distance to stop,$s=500m$

a=?

Use formula $v^{2}-u^{2}=2as$

$(0m/s)^{2}-(30m/s)^{2}=2*a*500m$

$0^{2}-30^{2}=2*a*500$

$-30^{2}/1000=a$

ANSWER: ${\color{Red} a=-0.9m/s^{2}}$

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Force=?

Use Formula $F=ma$

$F=750kg*0.9m/s^{2}$

ANSWER: ${\color{Red} F=675N}$

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B)

Horizontal Force,$F_{x}=500Ncos15 +800Ncos60$

$F_{x}=500cos15 +800cos60$

ANSWER: ${\color{Red} F_{x}=882.96N}$

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C)

(a)Held still

Tension= weight of crate =mg =50kg*9.81m/s²

ANSWER: ${\color{Red} T=490.5N}$

(b) If it is being lowered with an acceleration of 1.0 m/s²,

$T=mg-ma$

$T=m(g-a)$

$T=50kg(9.81m/s^{2}-1m/s^{2})$

$T=50*8.81$

ANSWER:${\color{Red} T=440.5N}$

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c)If it is being raised with an velocity of 1.0 m/s

Use Formula $T=mg+ma$

but since velocity is constant, acceleration is zero

$T=mg+m*0$

$T=mg$

$T=50kg*9.81m/s^{2}$

ANSWER: ${\color{Red} T=490.5N}$

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