This is a paradox. The answer is undefined.
The solution for this is:
Work done = force * distance = m*a*d and power = energy/time
The vo=0 and vf = 25 m/s and t=7 sec. This gives...
3.6 m/s^2 as acceleration and d=87.5 meters and thus F=ma= 5400 N.
Energy = 5400*87.5 = 4.7E5 Joules (2 sig. figs) and Power = 67,500 Watts or 90 HP (2 sig. figs again).
Answer:
27,000 m
450 m/s
Explanation:
Assuming the initial velocity is 0 m/s:
v₀ = 0 m/s
a = 15 m/s²
t = 60 s
A) Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (60 s) + ½ (15 m/s²) (60 s)²
Δy = 27,000 m
B) Find: v_avg
v_avg = Δy / t
v_avg = 27,000 m / 60 s
v_avg = 450 m/s
Answer:
(a) 81.54 N
(b) 570.75 J
(c) - 570.75 J
(d) 0 J, 0 J
(e) 0 J
Explanation:
mass of crate, m = 32 kg
distance, s = 7 m
coefficient of friction = 0.26
(a) As it is moving with constant velocity so the force applied is equal to the friction force.
F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N
(b) The work done on the crate
W = F x s = 81.54 x 7 = 570.75 J
(c) Work done by the friction
W' = - W = - 570.75 J
(d) Work done by the normal force
W'' = m g cos 90 = 0 J
Work done by the gravity
Wg = m g cos 90 = 0 J
(e) The total work done is
Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J
Answer: work = 1,305kJ
Explanation:
angle= 30°
force= 1,500N
distance= 1,000m
The formula for work is : Work= force x distance, however there is an angle of 30° between the direction of force applied and the direction of motion, therefore force must be decomposed to its value on the horizontal axis which is the direction of motion by using the cosine of the very angle.
W= F×cos(α)×D
W= 1,500×cos (30)×1,000
W= 1,305kJ ( kilojoules)
