Answer:
987 joules, 3.01s
Explanation:
(A)
from the attached diagram
net force, Fnet, pulling the crate up the ramp is given by
Fnet = FcosФ - WsinФ - Fr
where FcosФ is the component of horizontal force 290N resolved parallel to the plane
WsinФ = mgsinФ = component of the weight of the crate resolved parallel to the plane
Fr = constant opposing frictional force
Fnet = 290cos34⁰ - 20 × 9.8 × sin34° - 65
Fnet = 240.421 - 109.602 - 65
Fnet = 65.82N
Work done on the crate up the ramp, W, is given by
W = Fnet × d (distance up the plane)
W = 65.819 × 15
W = 987.285 joules
W = 987 joules (to 3 significant Figures)
(B)
to calculate the time of travel up the ramp
we use the equation of motion
[tex]s = ut + \frac{1}{2}at^{2}[/tex]
where s = distance up the plane, 15m
u = Initial velocity of the crate, which is 0 for a body that is initially at rest
a = acceleration up the plane, given by
[tex]a = \frac{Fnet}{m}[/tex]
where m = mass of the crate, 20 kg
now, [tex]a = \frac{65.819}{20} \\a = 3.291\frac{m^{2} }{s}[/tex]
from, [tex]s = ut + \frac{1}{2}at^{2}[/tex]
[tex]15 = 0*t + \frac{1}{2}* 3.291 * t^{2}[/tex]
15 = 0 + 1.645[tex]t^{2}[/tex]
15 = 1.645[tex]t^{2}[/tex]
[tex]t = \sqrt{\frac{15}{1.645} }[/tex]
[tex]t = 3.019[/tex]
t = 3.01s (to 3 sig fig)
(A) The total work done on the crate is 2.63 kJ
(B) The time taken to reach the top of the incline is 1.85 s.
According to the questions, the angle of the incline is θ = 34.0°, and the force F = 290N is acting horizontally. So it can be divided into 2 components as follows:
(i) force acting along the surface of the [tex]F_x=Fcos\theta[/tex]
(ii) force acting downward perpendicular to the inclined surface [tex]F_y=sin\theta[/tex]
(A) The frictional force acts opposite to the motion given by:
f = 65N, so the net force acting on the crate is ;
[tex]F' = Fcos\theta-f\\\\F'=290cos34^o-65\\\\F'=175.42N[/tex]
The work done is given by:
W = F'd
where d is the distance traveled
W = 175.42 × 15 J
W = 2.63 kJ
(B) the equation of motion of the crate along the surface of the incline is given by:
F' = ma
where m is the mass of the crate
a = F'/m
a = 175.42/20
a = 8.771 m/s
from the second equation of motion:
[tex]d = ut+\frac{1}{2}at^2[/tex]
here u is the initial speed = 0 m/s
and t is the time taken
[tex]t =\sqrt{\frac{2d}{a} }\\\\t=\sqrt{\frac{2\times30}{8.771} } \\\\t=1.85\;s[/tex]
Learn more about inclined surface:
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