1) 2063 N
The work done by a force is given by the equation:
[tex]W=Fd cos \theta[/tex]
where
F is the magnitude of the force
d is the displacement
[tex]\theta[/tex] is the angle between the direction of F and d
In this problem, we have
F = 157 N
d = 14.5 m
[tex]\theta=25^{\circ}[/tex]
Substituting,
[tex]W=(157)(14.5)(cos(25))=2063 N[/tex]
2) 12.7 m/s
The egg is in free fall, so it is moving by uniformly accelerated motion. This means that its final velocity can be found by using the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where we have:
v = final velocity
u = 0 initial velocity
[tex]a = g = 9.8 m/s^2[/tex] is the acceleration of gravity
s is the displacement
The building is 8.2 m high, so the egg falls through a distance of
d = 8.2 m
Therefore, substituting and solving the equation for v, we find:
[tex]v=\sqrt{u^2+2as}=\sqrt{0+(2)(9.8)(8.2)}=12.7 m/s[/tex]
3) 8.7 m/s
Assuming that no external forces acts on the Snookie-lemur system, then according to the law of conservation of momentum, their total momentum is conserved. This means that the initial momentum of Snookie in the wagon must be equal to the total momentum after the lemur appears in the wagon:
[tex]p_i = p_f\\M u = (M+m)v[/tex]
where
M = 215 kg is the mass of Snookie+the wagon
u = 9.5 m/s is the initial velocity of Snookie+the wagon
m = 19.7 kg is the mass of the lemur
v is the final velocity of Snookie+the wagon + the lemur
Solving the equation for v, we find:
[tex]v=\frac{Mu}{M+m}=\frac{(215)(9.5)}{215+19.7}=8.7 m/s[/tex]
4) Yes (he was travelling at 27.1 m/s)
According to the law of conservation of momentum, the total momentum of the two cars must be conserved before and after the collision.
This means that we can write the following equation:
[tex]m_B u_B + m_Z u_Z = (m_B + m_Z) v[/tex]
where:
[tex]m_B = 1100 kg[/tex] is the mass of Brady's car
[tex]m_Z = 1475 kg[/tex] is the mass of Jay Z car
[tex]u_B = 20 m/s[/tex] is the initial velocity of Brady's car
[tex]u_Z[/tex] is the initial velocity of Jay Z car
[tex]v=-7 m/s[/tex] is the final velocity of the two cars after they collide (in the negative direction, since it is in the same direction Jay Z was driving)
Solving for uZ, we find Jay Z initial velocity:
[tex]u_Z = \frac{(m_B+m_Z)v-m_B u_B}{m_Z}=\frac{(1100+1475)(-7)-(1100)(20)}{1475}=-27.1 m/s[/tex]
Therefore, Jay Z was driving faster than the limit of 25 m/s.
5) 166.6 N
The ball is moving in a uniform circular motion, so the tension in the string provides the centripetal force that keeps the ball in circular motion.
Therefore, we can write the following equation:
[tex]T=m\frac{v^2}{r}[/tex]
where
T is the tension
m is the mass of the ball
v is the speed of the ball
r is the radius of the circle
In this problem,
r = 0.23 m
v = 7.8 m/s
m = 0.63 kg
Substituting these numbers into the equation, we find the centripetal force, and therefore the tension in the string:
[tex]T=(0.63)\frac{7.8^2}{0.23}=166.6 N[/tex]
