Answer:
11 m/s
Explanation:
Draw a free body diagram. There are two forces acting on the car:
Weigh force mg pulling down
Normal force N pushing perpendicular to the incline
Sum the forces in the +y direction:
∑F = ma
N cos θ − mg = 0
N = mg / cos θ
Sum the forces in the radial (+x) direction:
∑F = ma
N sin θ = m v² / r
Substitute and solve for v:
(mg / cos θ) sin θ = m v² / r
g tan θ = v² / r
v = √(gr tan θ)
Plug in values:
v = √(9.8 m/s² × 48 m × tan 15°)
v = 11.2 m/s
Rounded to 2 significant figures, the maximum speed is 11 m/s.
Explanation:
It is given that,
Mass of the box, m = 100 kg
Left rope makes an angle of 20 degrees with the vertical, and the right rope makes an angle of 40 degrees.
From the attached figure, the x and y component of forces is given by :
Let 
and 
is the resultant in x and y direction.
As the system is balanced the net force acting on it is 0. So,
.............(1)
..................(2)
On solving equation (1) and (2) we get:
(tension on the left rope)
(tension on the right rope)
So, the tension on the right rope is 1063.36 N. Hence, this is the required solution.
Answer:
a. 4.9 m
Explanation:
To solve this problem we must take into account that power is defined as the relationship between the work and the time in which the work is done.
P = W/t
where:
P = power = 95 [W] (units of watts)
W = work [J] (units of Joules)
t = time = 6.2 [s]
We can clear the work from the previous equation.
W = P*t
W = 95*6.2 = 589 [J]
Now we know that the work is defined by the product of the force by the distance, therefore we can express the work done with the following equation.
W = F*d
where:
F = force = 120 [N] (units of Newtons)
d = distance [m]
d = W/F
d = 589/120
d = 4.9 [m]
