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I hope this helps ^-^
A mass suspended from a spring is oscillating up and down, (as stated but not indicated).
A). At some point during the oscillation the mass has zero velocity but its acceleration is non-zero (can be either positive or negative). <em>Yes. </em> This statement is true at the top and bottom ends of the motion.
B). At some point during the oscillation the mass has zero velocity and zero acceleration. No. If the mass is bouncing, this is never true. It only happens if the mass is hanging motionless on the spring.
C). At some point during the oscillation the mass has non-zero velocity (can be either positive or negative) but has zero acceleration. <em>Yes.</em> This is true as the bouncing mass passes through the "zero point" ... the point where the upward force of the stretched spring is equal to the weight of the mass. At that instant, the vertical forces on the mass are balanced, and the net vertical force is zero ... so there's no acceleration at that instant, because (as Newton informed us), A = F/m .
D). At all points during the oscillation the mass has non-zero velocity and has nonzero acceleration (either can be positive or negative). No. This can only happen if the mass is hanging lifeless from the spring. If it's bouncing, then It has zero velocity at the top and bottom extremes ... where acceleration is maximum ... and maximum velocity at the center of the swing ... where acceleration is zero.
The correct answer to the question is: 78.4 N.
CALCULATION:
As per the question, the mass of the object is given as m = 8 Kg.
We are asked to calculate the force of 8 Kg object pushing up from the table.
The force pushing up from the table is nothing else than the normal reaction.
The normal reaction of the object is equal to the weight of the object as the object is simply resting on the table.
Hence, normal reaction = Weight
= mg
= 8 × 9.8 N.
= 78.4 N.
Here, g is known as acceleration due to gravity.
Hence, the force of the object pushing up from the table is 78.4 N.
Answer:
m1 = 20g (= 0.02 kg)
Mass of pistol, m2 = 2 kg
Initial velocity of the bullet (u1) and pistol (u2) = 0
Final velocity of the bullet, v1 = +150m s-1
Let v be the recoil velocity of the pistol.
Total momentum of the pistol and bullet after it is fired is
= (0.02 kg x 150 m s-1) + (2 kg x v m s-1)
= (3 + 2v) kg m s-1
Total momentum after the fire = Total momentum before the fire
3 + 2v = 0
→v = -1.5 m/s
Answer:
v = √k/m x
Explanation:
We can solve this exercise using the energy conservation relationships
starting point. Fully compressed spring
Em₀ = 
= ½ k x²
final point. Cart after leaving the spring
= K = ½ m v²
Em₀ = Em_{f}
½ k x² = ½ m v²
v = √k/m x
