Question
Question: Find the values of \({F_1}\) and \({a_2}\) in the table given below. Mass \(m{\text{ (kg)}}\)| A...
Find the values of F1 and a2 in the table given below.
Mass m (kg) | Acceleration a (m/s2) | Force F (N) |
---|---|---|
25 | 1.2 | F1 |
1.5 | a2 | 2.25 |
A) 15, 1.2
B) 1.5, 20
C) 25, 1.5
D) 30, 1.5
Solution
Use Newton’ s second law of motion which gives force as the product of mass and acceleration to find F1 and a2
Formula Used: Force F acting on a body of mass mto provide an acceleration a to it is given by, F=ma
Complete step by step answer:
Step 1: List the information provided in the first row of the table
From the first row of the table we have,
Mass of the body, m1=25kg
Acceleration of the body, a1=1.2m/s2
Force F1 of the body is unknown
Step 2: Use the force equation F=ma to find F1
From the force equation we have F1=ma
Substituting the values of m1=25kg and a1=1.2m/s2 in the above equation
Then, we have F1=25×1.2=30N
i.e., the force applied on a body of mass of 25kg to produce an acceleration of 1.2m/s2 is 30N
Step 3: List the information provided in the second row of the table
From the second row of the table we have,
Mass of the body, m2=1.5kg
Force applied on the body, F2=2.25N
Acceleration a2of the body is unknown
Step 4: Use the force equation F=ma to find a2
From the force equation we have F2=m2a2
Expressing the force equation in terms of acceleration a2 we get, a2=m2F2
Substituting the values of m2=1.5kg and F2=2.25N in the above equation
Then, we have a2=1.52.25=1.5m/s2
i.e., when a force of 2.25N is applied on a body of mass of 25kg an acceleration of 1.5m/s2 is produced
Therefore, the correct option is d) 30, 1.5
Note: Newton’ s second law states that the rate of change of momentum (p) of a body is directly proportional to the applied force and takes place in the direction in which the force acts, i.e., F=dtdp
The momentum of the body is p=mv ,where m is the mass of the body and v is its velocity.
So Newton’ s second law can be stated as F=ma , where a is the body’s acceleration.