Pretty sure it’s Force*Distance*Cos(theta)
Answer:
W = Fd = KE =1/2mv²
Explanation:
not sure if that's what your looking for but i'm pretty sure this is it.
Answer:
a) k = 2231.40 N/m
b) v = 0.491 m/s
Explanation:
Let k be the spring force constant , x be the compression displacement of the spring and v be the speed of the box.
when the box encounters the spring, all the energy of the box is kinetic energy:
the energy relationship between the box and the spring is given by:
1/2(m)×(v^2) = 1/2(k)×(x^2)
(m)×(v^2) = (k)×(x^2)
a) (m)×(v^2) = (k)×(x^2)
k = [(m)×(v^2)]/(x^2)
k = [(3)×((1.8)^2)]/((6.6×10^-2)^2)
k = 2231.40 N/m
Therefore, the force spring constant is 2231.40 N/m
b) (m)×(v^2) = (k)×(x^2)
v^2 = [(k)(x^2)]/m
v = \sqrt{ [(k)(x^2)]/m}
v = \sqrt{ [(2231.40)((1.8×10^-2)^2)]/(3)}
= 0.491 m/s
Answer:
The new kinetic energy would be 16 times greater than before.
Explanation:
Kinetic energy is found using this formula:
- KE = 1/2mv²
- where KE = kinetic energy (J), m = mass (kg), and v = velocity (m/s)
We can see that kinetic energy is directly proportional to the square of the velocity, meaning that if the speed was increased by 4 times, then the kinetic energy would get increased by a factor of 16.
The velocity just before the ball hits the ground can be found by the equation:
Let's substitute h = 10 m and h = 40 m into this formula.
We can see that the velocity increases by a factor of 4 (10 m → 40 m).
Therefore, this means that the kinetic energy would also be increased by a factor of (4)² = 16. Thus, the answer is D) The new kinetic energy would be 16 times greater than before.