Answer:
The magnitude of the force is 0.7255kN
Explanation:
The elevator floor acts on the person with a force that is due to the gravitational acceleration less the downward acceleration of the elevator:
(force of floor F) = (mass of person m) x [ (grav. acceleration g) - (elevator acceleration a) ]
in other words, considering the elevator floor as a reference frame in the Earth's gravitational field, the person's weight decreases due to the downward acceleration, as follows:
We are given the person's weight at rest, 0.9kN, from which the mass can be determined as:
So
Answer:
Explanation:
Given that,
5J work is done by stretching a spring
e = 19cm = 0.19m
Assuming the spring is ideal, then we can apply Hooke's law
F = kx
To calculate k, we can apply the Workdone by a spring formula
W=∫F.dx
Since F=kx
W = ∫kx dx from x = 0 to x = 0.19
W = ½kx² from x = 0 to x = 0.19
W = ½k (0.19²-0²)
5 = ½k(0.0361-0)
5×2 = 0.0361k
Then, k = 10/0.0361
k = 277.008 N/m
The spring constant is 277.008N/m
Then, applying Hooke's law to find the applied force
F = kx
F = 277.008 × 0.19
F = 52.63 N
The applied force is 52.63N
Answer:
b. 20 sec
Explanation:
y = y₀ + v₀ t + ½ g t²
0 = 0 + (100) t + ½ (-10) t²
0 = 100t − 5t²
0 = t (100 − 5t)
t = 0, t = 20
The body lands after 20 seconds.
