First thing to do is to draw the system described above. Then, write an equation for the forces present.
<span>
</span>Σ<span>F = Fg - Ff
</span><span>0 = mgsin</span><span>∅</span><span> - umgcos</span><span>∅</span><span>0 = gsin</span><span>∅</span><span> - ugcos</span><span>∅</span><span>
u = tan</span><span>∅
</span>∅(max) = tan^-1 (u)<span>
</span>
Turn your protracted to a 90 degree angle
Answer:
Final velocity = 7.677 m/s
KE before crash = 202300 J
KE after crash = 182,702.62 J
Explanation:
We are given;
m1 = 1400 kg
m2 = 4700 kg
u1 = 17 m/s
u2 = 0 m/s
Using formula for inelastic collision, we have;
m1•u1 + m2•u2 = (m1 + m2)v
Where v is final velocity after collision.
Plugging in the relevant values;
(1400 × 17) + (4700 × 0) = (1400 + 1700)v
23800 = 3100v
v = 23800/3100
v = 7.677 m/s
Kinetic energy before crash = ½ × 1400 × 17² = 202300 J
Kinetic energy after crash = ½(1400 + 1700) × 7.677² = 182,702.62 J
The only vertical forces are weight and normal force, and they balance since the surface is horizontal. The horizontal forces are the applied force (uppercase F) in the direction the block slides and the frictional force (lowercase f) in the opposite direction.
Apply Newton's 2nd Law in the horizontal direction:
ΣF = ma
F - f = ma
where f = µmg
F - µmg = ma
F = m(a +µg)
F = (20 kg)(1.4 m/s² + 0.28(9.8 m/s²)
F = 83 N