Answer:
A)
0.395 m
B)
2.4 m/s
Explanation:
A)
= mass of the cart = 1.4 kg
= spring constant of the spring = 50 Nm⁻¹
= initial position of spring from equilibrium position = 0.21 m
= initial speed of the cart = 2.0 ms⁻¹
= amplitude of the oscillation = ?
Using conservation of energy
Final spring energy = initial kinetic energy + initial spring energy
B)
= mass of the cart = 1.4 kg
= spring constant of the spring = 50 Nm⁻¹
= amplitude of the oscillation = 0.395 m
= maximum speed at the equilibrium position
Using conservation of energy
Kinetic energy at equilibrium position = maximum spring potential energy at extreme stretch of the spring
Answer:
a) 4.31 m/s²
b) 215.5 m
Explanation:
a) According to Newton's first law of motion
The net force applied to particular mass produced acceleration, a, according to
F = ma
F = 140 N
m = 32.5 kg
a = ?
140 = 32.5 × a
a = 140/32.5 = 4.31 m/s²
b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s
u = initial velocity of the probe = 0 m/s (since it was initially at rest)
a = 4.31 m/s²
t = 10 s
s = distance travelled = ?
s = ut + at²/2
s = 0 + (4.31×10²)/2 = 215.5 m
A contact force is a type of force which act on an object by coming in contact with the object. Examples of contact force that acts through a force field are: applied force, frictional force, air resistance force, tension, spring force, etc.
Examples of forces that act through a force field are gravitational force, electromagnetic force, the weak interaction and the strong interaction.
