Newton's Second Law of Motion Question?

Question

Hi,
Here is the question I am trying to solve:
According to Newton's Second law of motion, the force F acting upon an object varies directly with its acceleration a. A force of 63N acting upon a given object results in an acceleration of 9 m/s^2. If the magnitude of the force acting upon the same object is changed to 28 N, what will the object's acceleration become?

I have been trying out how to figure the above problem for a bit, but the formulas confuse me. X_X

I really appreciate your help!

Answer 1

a = 4m / s^2

Answer 2

Force = mass * acceleration

You can rearrange that equation however you need to to solve for the missing parameter.

In this case, you know the initial Force (63 N) and the acceleration (9 m/s^2). You put the knowns on the same side of the equation:

(63 N) / (9 m/sec^2) = mass = 7 kg

Now you have a new force on the same object (mass stays constant), so you have to rearrange the F=ma equation to put your knowns on the same side of the equation:

(28 N)/ (7 kg) = acceleration = 4 m/sec^2

If you're doing dimensional analysis, it would help to know that 1 N = (1 kg m)/sec^2. It helps as a double check to make sure you did your algebra correctly when you're moving your knowns to the same side of the equation.

Answer 3

well ... newton's second law states that the force is equal to the rate of cange of momentum
Momentum(M) is equal to the mass of the object multiplied by its velocity
So M = mass * velocity
and so .. Force = (mass * velocity ) / time .. - as force is equal to the rate of change of momentum-
and so we could say that Force = mass * (velocity/time)
and since that the velocity/time is the acceleration
We deduce that Force= mass * acceleration


considering the problem u r trying to solve
Force =63 N making acceleration = 9 m/s^2
by using the the equation F=ma
therefore mass of the object is 7 kg (63/9)
mass=7 kg
and so when the force on the body changes to 28N
therefore the acceleration of the object will be 4 m/s^2 (28/7)
again using the equation F=ma

Answer 4

F = m*a

Force = mass * acceleration

If the force is n times smaller, the acceleration will also be n times smaller F ~ a

In this case the objects acceleration will be 4 m/s^2

a=(28/63)*9 = 4 m/s^2

Answer 5

you know F = 63 and a = 9, then

F = m * a

shows that

m = F/a = 63 / 9 = 7 kg

The 28 N force leads to an acceleration of

a = F/m = 28 / 7 = 4 m/s^2

Answer 6

F=m*a

F=force
m=mass
a=acceleration

A newton ( N) = kg*m/s^2 where the 'm' here refers to meter, not mass.

F and a are given, solve for m.
m = F/a = 63 (kg*m/s^2)/9(m/s^2) = 7kg

In the second part of the problem, force is given and you have the mass from the first part of the problem. Solve for acceleration.

F=m*a
a=F/m=28N/7kg=4m/s^2

Answer 7

as F=m*a , 63=m*9.
so m=7kg. the mass of the object will remain constant.
so 28=7*a. therefor a=28/7 =4 m/s2

Answer 8

since the variation is direct
(28/63)*9=4 m/s/s

Answer 9

Force = mass * acceleration

Mass stays constant, so:

F1 = m * a1 -----> 63 [kg*m/sec^2] = m * 9 [m/sec^2]

F2 = m * a2 -----> 28 [kg*m/sec^2] = m * A

63/28 = 9 [m/sec^2] / A
A = 9*(28/63) [m/sec^2]
A = 4 [m/sec^2]

Answer 10

the problem is probably because the acceleration is 9.8 m/s/s, not just 9