Answer:
a. 
Explanation:
The equation of the forces along the directions parallel and perpendicular to the slope are:
- Along the parallel direction:
where
:
m = 6.0 kg is the mass of the box
g = 9.8 m/s^2 the acceleration of gravity
is the angle of the slope
is the coefficient of friction
R is the normal reaction
a is the acceleration
- Along the perpendicular direction:
From the 2nd equation, we get an expression for the reaction force:
And substituting into the 1st equation, we can find the acceleration:
Solving for a,
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Let N be the normal force that forces the person against the wall.
Then u N = m g is the frictional force supporting the person's weight
and N = m g / u
also, N = m v^2 / R is the normal force providing the centripetal acceleration
So, m g / u = m v^2 / R
v^2 = g R / u
since v = 2 pi R T
4 pi^2 R^2 T^2 = g R / u and T^2 = g / (4 u pi^2 R)
T = 1/ (2 pi) (g /(u R))^1/2 = .159 * (9.8 m/s^2 / (.521 * 4.4 m)) ^1/2
T = .68 / s
Do you see any thing wrong here?
T should have units of seconds not 1 / seconds
v should be 2 * pi * R / T where T is the time for 1 revolution
So you need to make that correction in the above formula for v.
An interesting problem, and thanks to the precise heading you put for the question.
We will assume zero air resistance.
We further assume that the angle with vertical is t=53.13 degrees, corresponding to sin(t)=0.8, and therefore cos(t)=0.6.
Given:
angle with vertical, t = 53.13 degrees
sin(t)=0.8; cos(t)=0.6;
air-borne time, T = 20 seconds
initial height, y0 = 800 m
Assume g = -9.81 m/s^2
initial velocity, v m/s (to be determined)
Solution:
(i) Determine initial velocity, v.
initial vertical velocity, vy = vsin(t)=0.8v
Using kinematics equation,
S(T)=800+(vy)T+(1/2)aT^2 ....(1)
Where S is height measured from ground.
substitute values in (1): S(20)=800+(0.8v)T+(-9.81)T^2 =>
v=((1/2)9.81(20^2)-800)/(0.8(20))=72.625 m/s for T=20 s
(ii) maximum height attained by the bomb
Differentiate (1) with respect to T, and equate to zero to find maximum
dS/dt=(vy)+aT=0 =>
Tmax=-(vy)/a = -0.8*72.625/(-9.81)= 5.9225 s
Maximum height,
Smax
=S(5.9225)
=800+(0.8*122.625)*(5.9225)+(1/2)(-9.81)(5.9225^2)
= 972.0494 m
(iii) Horizontal distance travelled by the bomb while air-borne
Horizontal velocity = vx = vcos(t) = 0.6v = 43.575 m/s
Horizontal distace travelled, Sx = (vx)T = 43.575*20 = 871.5 m
(iv) Velocity of the bomb when it strikes ground
vertical velocity with respect to time
V(T) =vy+aT...................(2)
Substitute values, vy=58.1 m/s, a=-9.81 m/s^2
V(T) = 58.130 + (-9.81)T =>
V(20)=58.130-(9.81)(20) = -138.1 m/s (vertical velocity at strike)
vx = 43.575 m/s (horizontal at strike)
resultant velocity = sqrt(43.575^2+(-138.1)^2) = 144.812 m/s (magnitude)
in direction theta = atan(43.575,138.1)
= 17.5 degrees with the vertical, downward and forward. (direction)
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I hope some of this information I gave you can help you. I came up with everything myself to help you.
