Question

solve question no. 16,19and 20​

Answer 1

Answer:

16. (3) 4 Newton

19. (1) 100 Newton

20. (4) 0 Newton

Explanation:

16. If both blocks are moving together, then acceleration of the 2 kg block would also be 2 m/s² as shown in the figure.

Now, ∵ F = m × a

∴ Force = 2 kg × 2 m/s² = 4 Newton.

19. If breaking strength of string is 60 Newtons, then a maximum of 60 Newton of force can be applied on the 3 kg block.

Hence, acceleration of 3 kg block when maximum force is acting will be

$a_{max}={F_{max}\over m}={60\over 3}=20~m/s^2$

Now, the value of acceleration will be same for both the blocks, because the 2 kg block is pulling the latter one. So,

Value of maximum force that can be applied on the system of (2 kg + 3 kg) block will be

$F_{max}=mass\times a_{max}=5\times 20=100~Newton.$

20. I can't get what this question is saying, because without decelerating or accelerating, the car can't take any turn. Also, it hasn't mentioned any circular motion.

As no acceleration or retardation is mentioned, and the velocity given is constant, then there must be no force acting on the body.

Answer 2

16) mass = 2+4 = 6kg

acceleration = 2m/s²

force= mass ×acceleration

force = 6× 2 = 12N

19)