Answer:
Kathmandu
Explanation:
As the altitude get higher, the gravitational pull of the earth on the object increases, therefore, the mass is higher up above.
1) draw a diagram.
2) label diagram. (split the 100 degrees into 50, (which is right down the middle) to make a right angle triangle.)
3) since its a free body diagram, the forces known must be labelled. (force of gravity). this shows that the straight vertical line of the right angle triangle is Fg (force gravity). label it.
4) use trigonometry. rearrange the equation to solve for what needs to be known.
angles known: 50 (split 100 in half to make a right angle triangle), 90 (since its right angle), and 40 (180-90-50 = 40)
sides known: vertical lined up with the 90 degree angle. Fg. --> fg=mg=500N x 9.81m/s^2 = 4905N
use formula: sin or cos
i used sin. sin(40) = 4905 / ?
- times '?' on both sides. : sin(40) x '?' = 4905
-divide both sides by sin(40): '?' = 4905/ sin(40)
--> Solve.
W=Fd. Force is not given so we solve for it. F=ma, m=5kg, a=2m/s^2, F=10N. Distance is not given so we solve for it, x=.5a(t^2)=.5(2)(7x7)=49m. F=10N, d=49m, W=490J.
Answer:
a) v = √ g x
, b) W = 2 m g d
, c) a = ½ g
Explanation:
a) For this exercise we use Newton's second law, suppose that the block of mass m moves up
T-W₁ = m a
W₃ - T = M a
w₃ - w₁ = (m + M) a
a = (3m - m) / (m + 3m) g
a = 2/4 g
a = ½ g
the speed of the blocks is
v² = v₀² + 2 ½ g x
v = √ g x
b) Work is a scalar, therefore an additive quantity
light block s
W₁ = -W d = - mg d
3m heavy block
W₂ = W d = 3m g d
the total work is
W = W₁ + W₂
W = 2 m g d
c) in the center of mass all external forces are applied, they relate it is
a = ½ g
