From the graph, it can be seen that the constant force that John exerted in order to move the object is 14N. Work is calculated by multiplying the force with the distance to which the object moves in parallel with the direction of the force.
Work = Force x displacement
Work = (14 N) x (8 m)
Work = 112 J
The closest value is 110J. Thus, the answer to this item is the second choice.
The equation of the car is given by the equation,
x(t) = 2.31 + 4.90t² - 0.10t⁶
If we are going to differentiate the equation in terms of x, we get the value for velocity.
dx/dt = 9.8t - 0.6t⁵
Calculate for the value of t when dx/dt = 0.
dx/dt = 0 = (9.8 - 0.6t⁴)(t)
The values of t from the equation is approximately equal to 0 and 2.
If we substitute these values to the equation for displacement,
(0) , x = 2.31 + 4.90(0²) - 0.1(0⁶) = 2.31
(2) , x = 2.31 + 4.90(2²) - 0.1(2⁶) = 15.51
Thus, the positions at the instants where velocity is zero are 2.31 and 15.51 meters.
-- In order to achieve constant verlocity, the net force on the mass must be zero. So if there ARE any forces acting on it, they must be balanced.
-- There is already a force on the mass that can't be eliminated . . . the force of gravity.
-- That force due to gravity is (mass x gravity) = (25 kg)(9.8 m/s²) = <em><u>245N</u></em> in the <u><em>downward</em></u> direction.
-- In order to 'balance' the forces and make them add up to zero, we have to provide another force of <em>245N</em>, all in the <em>upward</em> direction.
-- Then the forces on the object will be balanced, the NET force on it will be zero, and whichever way you start it moving, it will continue to move at a cornstant verlocity.
Answer:
1.73 seconds
Explanation:
The velocity the ball first hits the ground with is:
v² = v₀² + 2aΔx
v² = (0 m/s)² + 2 (-10 m/s²) (-20 m)
v = -20 m/s
The velocity it rebounds with is 3/4 of that in the opposite direction, or 15 m/s.
The time it takes to return to the ground is:
Δx = v₀ t + ½ at²
0 = (15 m/s) t + ½ (-10 m/s²) t²
0 = t (15 − 5t²)
t = √3
t ≈ 1.73 seconds
