A constant retarding force of 50 \(\text N\) is applied to a body of mass 20 \(\text {kg}\) moving initially with a speed of 15 \(\text m \,\text s^{-1}\). How long does the body take to stop ?
Solution and Explanation
Retarding force, \(\text F\) = –50 \(\text N\)
Mass of the body, \(\text m\) = 20 \(\text {kg}\)
Initial velocity of the body, \(\text u\) = 15 \(\text m/\text s\)
Final velocity of the body, \(\text v\) = 0
Using Newton’s second law of motion, the acceleration (a) produced in the body can be calculated as:
\(\text F\) = \(\text {ma}\)
–50 = 20 × \(\text a\)
\(\therefore\) \(\text a\) = \(\frac{-50}{20}\) = -2.5 \(\text m/\text s^2\)
Using the first equation of motion, the time (t) taken by the body to come to rest can be calculated as:
\(\text v\) = \(\text u+\text {at}\)
∴ \(\text t\) = \(\frac{-u}{a}\) = \(\frac{-15}{-2.5}\) = 6 \(\text s\)
Top Questions on Newton’s Second Law Of Motion
- A mass of 200 gm has initial velocity \(v_i = 2\hat{i}+3\hat{j}\) and final velocity \(v_f=-2\hat{i}-3\hat{j}\) find magnitude of change in momentum
- JIPMER MBBS - 2018
- Physics
- Newton’s Second Law Of Motion
- A stone of 1 kg is thrown with a velocity of 20 m s–1 across the frozen surface of a lake and comes to rest after traveling a distance of 50 m. What is the force of friction between the stone and the ice?
- CBSE Class IX
- Science
- Newton’s Second Law Of Motion
- How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s–2.
- CBSE Class IX
- Science
- Newton’s Second Law Of Motion
- An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m s–2?
- CBSE Class IX
- Science
- Newton’s Second Law Of Motion
- Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
- CBSE Class IX
- Science
- Newton’s Second Law Of Motion
Questions Asked in CBSE Class XI exam
- We have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
- CBSE Class XI
- Motion in a straight line
- If U = {a, b, c, d, e, f, g, h}, find the complements of the following sets:
(i) A = {a, b, c}
(ii) B = {d, e, f, g}
(iii) C = {a, c, e, g}
(iv) D = {f, g, h, a}- CBSE Class XI
- Complement of a Set
- What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?
(a) KI3 (b) H2S4O6 (c) Fe3O4 (d) CH3CH2OH (e) CH3COOH- CBSE Class XI
- Oxidation Number
- Using s, p, d notations, describe the orbital with the following quantum numbers.
- n=1, l=0
- n = 3, l=1
- n = 4, l =2
- n=4, l=3
- CBSE Class XI
- Developments Leading to the Bohr’s Model of Atom
- Write the following as intervals:
(i) {\(x: x ∈ R, -4 < x ≤ 6\)}
(ii) {\(x: x ∈ R, -12 < x < -10\)}
(iii) {\(x: x ∈ R, 0 ≤ x < 7\)}
(iv) {\(x: x ∈ R, 3 ≤ x ≤ 4\)}- CBSE Class XI
- Sets and their Representations
Concepts Used:
Newtons Laws of Motion
Newton’s First Law of Motion:
Newton’s 1st law states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it.
Newton’s Second Law of Motion:
Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object’s mass.
Mathematically, we express the second law of motion as follows:
Newton’s Third Law of Motion:
Newton’s 3rd law states that there is an equal and opposite reaction for every action.