5.6 A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s-1 to 3.5 m s-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
5.6 A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s-1 to 3.5 m s-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
-
1 Answer
-
Initial speed, u = 2.0 m/s, the final speed v = 3.5 m/s, the time t = 25 s
From the relation v = u + at, we get acceleration a = (v-u)/t = (3.5 – 2.0 )/ 25s = 0.06 m/s2
The force F = ma, F = 3 0.06 N = 0.18 N
Similar Questions for you
= 0.92 * 1260 = 1161 m/s
For 2 kg block
T – 2g sin37 = 2a . (i)
For 4 kg block
4g – 2T =
2g – T = a . (ii)
T = (2g – a)
2g – a – 2g × = 2a
3a = 2g ×
8m1 – 8m2 = m1 + m2
7m1 = 9m2
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 687k Reviews
- 1800k Answers