Assuming it is on a horizontal surface:
friction = μR
R = 20g (g is gravity 9.81)
so Friction = 0.085 x 20g
Work done is force x distance
so Work done = 0.085 x 20g x 28
= 466.956 J
Answer:
659.01W
Explanation:
The cab has a mass of 1250 kg, the weight of the cab represented by Wc will be
Wc = mass of the cab × acceleration due to gravity in m/s²
Wc = 1250 × 9.81 = 12262.5 N
but the counter weight of the elevator represented by We = mass × acceleration due to gravity = 995 × 9.81 = 9760.95 N
Net weight = weight of the cab - counter weight of the elevator = Wc - We = 12262.5 - 9760.95 = 2501.55 N
the motor of the elevator will have to provide this in form of work
work done by the elevator to lift the cab to height of 49 m = net weight × distance (height) = 2501.55 × 49m
power provided by the motor of the elevator = workdone by the motor / time in seconds
Power = (2501.55 × 49) ÷ ( 3.1 × 60 seconds) = 659.01 W
Answer:
a) the distances are zero, Both 1st & 2nd condition
c) the torques are equal but of the opposite sign, 2nd condition of equilibrium
Explanation:
The equilibrium conditions are
1 translational
∑ F = 0
2 rotational
∑ τ = Σ (F_i x r_i) = 0
They tell us that external torque is zero.
Therefore we have two various possibilities
a) the distances are zero, in this case we have a pure translation movement
for this situation the two equilibrium relations are fulfilled
b) the forces are zero, there is no movement
It does not make sense to use the equilibrium relations since there are no forces
c) the torques are equal but of the opposite sign, the forces are on the opposite side of the body.
In this case the 2 equilibrium relation is fulfilled, but not the first one that the force has the same direction
Answer:
1) 10.1 s 2) 909 m 3) 90.0 m/s 4) -99m/s 5) just over the bomb.
Explanation:
1)
- In the vertical direction, as the bomb is dropped, its initial velocity is 0.
- So, we can find the time required for the bomb to reach the earth, applying the following kinematic equation for displacement:
- where Δy = -500 m (taking the upward direction as positive).
- a=-g=-9.8 m/s²
- Replacing these values in (1), and solving for t, we have:
- The time required for the bomb to reach the earth is 10.1 s.
2)
- In the horizontal direction, once released from the helicopter, no external influence acts on the bomb, so it will continue moving forward at the same speed. that it had, equal to the helicopter.
- As the time must be the same for both movements, we can find the horizontal displacement just as the product of this speed times the time, as follows:
3)
- The horizontal component of the bomb's velocity is the same that it had when left the helicopter. i.e. 90 m/s.
4)
- In order to find the vertical component of the bomb's velocity just before it strikes the earth, we can apply the definition of acceleration, remembering that v₀ = 0, as follows:
5)
- If the helicopter keeps flying horizontally at the same speed, it will be always over the bomb, as both travel horizontally at the same speed.
- So, when the bomb hits the ground, the helicopter will be exactly over it.
