Polly is pushing a box across the floor with a force of 30 N. The force of gravity is -8 in, and the normal force is eight in. W
2 answers:
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Answer:
a)3.5s
b)28.57m/S
c)34.33m/S
d)44.66m/S
Explanation:
Hello!
we will solve this exercise numeral by numeral
a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration
where
Vo = Initial speed =0
T = time
g=gravity=9.81m/s^2
y = height=60m
solving for time
T=3.5s
b)The horizontal speed remains constant since there is no horizontal acceleration. with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed
c)
to find the final vertical velocity we use the equations for motion with constant velocity as follows
Vf=Vo+g.t
Vf=0+(9.81 )(3.5)=34.335m/S
d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares
Answer:
W=561.41 J
Explanation:
Given that
m = 51 kg
μk = 0.12
θ = 36.9∘
Lets F is the force applied by man
Given that block is moving at constant speed it mans that acceleration is zero.
Horizontal force = F cos θ
Vertical force = F sinθ
Friction force Fr= μk N
N + F sinθ = m g
N = m g - F sinθ
Fr = μk (m g - F sinθ)
For equilibrium
F cos θ = μk (m g - F sinθ)
F ( cos θ +μk sinθ) = μk (m g
Now by putting the values
F ( cos 36.9∘ + 0.12 x sin36.9∘)=0.12 x 51 x 10
F= 70.2 N
We know that Work
W= F cos θ .d
W= 70.2 x cos 36.9∘ x 10
W=561.41 J
