The potential energy would be zero. Only kinetic energy is present in this case. To find out what the answer is we do the equation: mv^2/2 soo...
KE =mv^2/2
KE= 1(2^2)/2 which the answer will come up by 2 Joules.
Answer:
The horizontal component of the velocity is 188 m/s
The vertical component of the velocity is 50 m/s.
Explanation:
Hi there!
Please, see the figure for a graphic description of the problem. Notice that the x-component of the vector velocity (vx), the y-component (vy) and the vector velocity form a right triangle. Then, we can use trigonometry to obtain the magnitude of vx and vy:
We can find vx using the following trigonometric rule of a right triangle:
cos α = adjacent / hypotenuse
cos 15° = vx / 195 m/s
195 m/s · cos 15° = vx
vx = 188 m/s
The horizontal component of the velocity is 188 m/s
To calculate the y-component we will use the following trigonometric rule:
sin α = opposite / hypotenuse
sin 15° = vy / 195 m/s
195 m/s · sin 15° = vy
vy = 50 m/s
The vertical component of the velocity is 50 m/s.
Answer:
A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s.
how much work is done on the monitor by (a) friction, (b) gravity
work(friction) = 453.5J
work(gravity) = -453.5J
Explanation:
Given that,
mass = 14kg
displacement length = 5.50m
displacement angle = 36.9°
velocity = 2.30cm/s
F = ma
work(friction) = mgsinθ .displacement
= (14) (9.81) (5.5sin36.9°)
= 453.5J
work(gravity)
= the influence of gravity oppose the motion of the box and can be pushing down, on the box from and angle of (36.9° + 90°)
= 126.9°
work(gravity) = (14) (9.81) (5.5cos126.9°)
= -453.5J