Problems #3

Examples and problems to Dynamics
Examples
1. a) A block is projected up along an inclined plane where the angle of the plane α = 35o
and the its length is l = 8 m. The coefficient of sliding friction is µ = 0,3. Find the
initial speed, to reach the block’s traveling just up to the top of the plane. b) What will
be the final speed of the block at the bottom of the plane, when it slides back?
Solution:
I.
a)
Fn
a
Gsinα
Ff
Gcosα
α
G
Using v 2 = v02 + 2as, 0 = v02 + 2al
From the components of the forces along the plane:
ma = −mg sin α − F f
The magnitude of frictional force is F f = µ Fn = µ mg cos α ,
since the net force perpendicular to the plane must be zero.
Using the previous equations
a = − g sin α − µ g cos α = −8,193m / s 2
Using this acceleration, the initial speed: v0 = 11,45 m/s
b) Taking down the plane as the positive direction in the next figure
Fn
Ff
a1
Gsinα
Gcosα
α
G
The acceleration comes from the different forms of the previous equations:
v 2 = 2 a1l
ma1 = mg sin α − F f
Hence
and the final speed
a1 = g sin α − µg cos α = 3,28m / s 2
v = 7,24 m/s
II.
a) In this solution of the same problem we use the terms of energies.
1
1
Using the work-energy theorem: W = mv 2 − mv02
2
2
1 2
− mgh − F f l = 0 − mv0 ; h = l sin α
2
Hence we will get the same result.
b) When the block slides down along the plane, the form of the work-energy theorem:
1
1 2
mv − 0
2
The result will be the same, as in the I. solution.
mgh − F f l =
2. a) A body with mass m = 0,5 kg moves in a straight line. The applied force as a
function of time is given by F = 3[N ] + 4[N / s ]t . The time interval of its motion is
t = 0,1 s. Find the velocity and acceleration of the body at the end of this interval.
b) Consider the acting force as a function of x, along the x-axis: F = 3[N ] + 4[N / m] x .
Find the velocity and acceleration at x = 0,05m from the starting point.
Solution:
a) Using the Newton’s second law
F 3 + 4t
a= =
= 6,8m / s 2
m
m
For the calculation the magnitude of velocity we can use the impulse-momentum theorem:
t
dI
F=
Hence ∫ F dt = mv
dt
0
That is the impulse is equal to the linear momentum at t time (vo = 0).
t
t2
∫0 (3 + 4t )dt = 3t + 4 2 = mv
Substituting the data
v = 0,64 m/s.
b) Using the Newton’s second law
F 3 + 4x
a= =
= 6,4m / s 2
m
m
Now the speed can be find from the work-energy theorem:
x
1 2
∫0 F dx = 2 mv
x
x
∫ (3 + 4 x )dx = 3x + 4
2
0
Hence
2
=
1 2
mv
2
v = 0,79 m/s.
Problems
1. A waitress shoves a ketchup bottle with mass 0,48 kg to the right along a smooth counter
with speed of 2,8 m/s. The bottle slows down as it slides because of constant horizontal
frictional force exerted on it by the counter top. It comes to rest after moving 1 m. What
are the magnitude and direction of the frictional force acting on the bottle? (Ff = - 1,88 N)
2. A stone of mass 0,2 kg slides down the smooth surface of the roof shown in the figure.
The initial velocity is zero. (α= 40o, g = 10 m/s2)
a) When will it reach the edge of the roof? What is the speed of the stone at leaving the
roof? (t1 = 1,21 s; v1 = 7,75 m/s)
b) What is the magnitude of velocity when the stone strikes the ground? (v = 18,97 m/s)
c) Find the final velocity, if there is a friction on the roof, and the coefficient of kinetic
friction is 0, 4. (v’ = 18,2 m/s)
2
3m
α
15m
3. A 4,8 kg bucket of water is accelerated upward by a cord of negligible mass whose
breaking strength is 75 N. Find the maximum upward acceleration that can be given to the
bucket without breaking the cord. (g = 10 m/s2) (a = 5,625 m/s2)
4. You try to move a 50 kg crate by tying a rope around it and pulling upward the rope at an
angle 30o above the horizontal. How hard you have to pull to keep the crate moving with
constant velocity? The coefficient of kinetic friction is 0,4. Is this easier, or harder than
pulling horizontally? (F = 187,61 N; horizontally F = 200 N)
5. In the previous problem suppose you try to move the crate with 210 N (the conditions are
the same). What will be the speed after 6 m, if the magnitude of initial velocity is 1 m/s?
(v=2,59 m/s)
6. The position of body with m mass is given in function time by x(t)=At-Bt3 where A and B
are constant. Find the force as a function of time. ( F (t ) = ma = −6mBt )
7. We want to load a 12 kg crate into a truck by sliding it up a ramp, inclined at 30o, giving it
an initial speed of 5 m/s at the bottom and letting it go. The crate slides 1,6 m up the ramp,
stops and slides back down. Find the kinetic frictional force. (Ff = 33,75 N)
v0
α
8. A 4 kg ball is thrown up at an angle to the horizontal. At the top of its trajectory, at 6 m
height it will stop after 2 m on the top of a wall with horizontal surface. The frictional
force acting on it is 5 N. What is the magnitude of the initial velocity? What is the initial
distance of the ball from the wall? (vo = 11,18 m/s; x = 2,45 m)
9. The power of a car is 22 kW, its speed is 100 km/h. Find the force acting by the engine of
the car. (F = 790 N)
10. Find the average power of the 2000 kg car, if it accelerates from rest to 108 km/h in 10 s.
(P = 45 kW)
3

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