Question

rack of mass 1500 kg moving at 30 m/s is braked suddenly wit 1000-V. How far does the car travel before stopping? th a constant braking force of 10. A baseball batter hits an incoming 45 m/s fastball. The ball leaves the bat at 55 m/s after a ball-on-bat contact time of 0.025 s. What is the force exerted on a 0.15-kg baseball? (choose + direction) 11. A trapeze artist, with swing, weighs 800 N; he is momentarily held horizontal by his partner so that swing ropes make an angle of 2.0° with the vertical. In such a condition of static equilibrium, what is the horizontal force being applied by the partner? Label the 3 forces on the diagram 20° 12. Complete the free-body diagram, Show all forces. Calculate the acceleration of the system shown. The horizontal surface has a frictional force of 100 N that acts in the negative direction. 30" 100kgx 13. A box of mass 10 kg is at rest on a horizontal frictionless surface. A constant horlzontal force F then acts on the box, and accelerates it to the right. It is observed that it takes the box 6.0 seconds to travel 30 meters. What is the magnitude of the force F? 10kg 14. A 50 kg mass is on a frictionless inclined plane shown in the diagram. Calculate the force, F, required to hold the mass in place on an inclined plane Inslude a free-body diagram. 30°

Answer 1

9. v_f² - v_i² = 2ax

v_f = final speed = 0 (full stop)

v_i = initial speed = 30 m/s

x = distance traveled

a = acceleration = F/m = force/mass

F = -10000 N (negative sign it it is opposing the direction of motion)

a = F/m = -10000 N / (1500 kg) = -6.67 m/s²

0 - v_i² = 2ax

x = -v_i² / 2a = - (30²) / 2(-6.67)

x = 67.47 m

10. using the Impulse-Momentum Theorem, which states that the magnitude of the impulse exerted on an object is equal to the magnitude of the change in momentum:

$F\, \Delta t=m\, \Delta v$

$\Rightarrow F=\frac{m\, \Delta v}{ \Delta t}$

$\Rightarrow F=\frac{m\, (v_{f}-V_{i})}{ \Delta t}$

$\Rightarrow F=\frac{(0.15)\, (55-(-45))}{ 0.025}$

$\therefore F=600\;N$

$11.\;tan\theta =\frac{opp}{adj}$

  $\Rightarrow opp=(adj)(tan\theta )$

  $\Rightarrow F=(800N)\tan 20^{o}$

$\therefore F=291.18\;N$