1)A constant force is applied to an object, causing the object to accelerate at

Question

1)A constant force is applied to an object, causing the object to accelerate at 14.0 m/s2 . 1.What will the acceleration be if the force is halved? 2.What will the acceleration be if the object's mass is halved? 3.What will the acceleration be if the force and the object's mass are both halved? 4.What will the acceleration be if the force is halved and the object's mass is doubled?

2)For an object starting from rest and accelerating with constant acceleration, distance traveled is proportional to the square of the time.

If an object travels 1.7 furlongs in the first 2.0 s, how far will it travel in the first 4.0 s?

Explanation / Answer

1) Given: A constant force is acting on an object; Object is accelerating at a = 14 m/s2

Solution:

1. Force is halved

The mass of the object, m shall remain constant hence

As F / 2 = ma / 2 [Using Newton's 2nd law, F = ma]

The acceleration gets halved, a = 7 m/s2

2. Object's mass is halved

As the force is constant, the object's mass being halved will cause the acceleration to be increased by a factor n to keep the force constant

F = (m/2)* na = ma

n*ma = 2*ma

n = 2

So, acceleration will be doubled, a = 28 m/s2

3. Force and mass both are halved

If both force and mass are halved then the equation becomes

F / 2 = ma / 2 => F = ma

So, acceleration remains the same, a = 14 m/s2

4. Force is halved and object's mass is doubled

If force is halved then

F / 2 = ma/2 => F' = m(a/2)

If mass is doubled, then acceleration will change by a factor n to compensate

F' = 2m(na/2) = m(a/2)

2n = 1

n = 1/2

F' = 2m(a/4)

So, acceleration will be 1/4th of its value, a = 3.5 m/s2

2) Given: Distance travelled in first 2s (t1) , d1 = 1.7 furlongs

To Find: Distance travelled in first 4s (t2) , d2

Solution:

d1 / d2 = t12 / t22

d2 = d1 * t22 / t12

d2 = 1.7 * 42 / 22 = 1.7 * 4

d2 = 6.8 furlongs

Ans. The object travels 6.8 furlongs

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