Potential Energy (Initial one) = m * g * h
P.E. = 60 * 9.8 * 10
P.E. = 5880
Kinetic Energy (Final One) = 1/2 mv²
K.E. = 1/2 * 60 * (10)²
K.E. = 6000/2
K.E. = 3000
Lost Energy = 5880 - 3000 = 2880 J
In short, Your Answer would be 2880 Joules
Hope this helps!
Answer:
No, distance is more important.
Answer:
The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.
Explanation:
Given:-
- The diameter of the drill bit, d = 98 cm
- The power at which drill works, P = 5.85 hp
- The rotational speed of drill, N = 1900 rpm
Find:-
What Torque And Force Is Applied To The Drill Bit?
Solution:-
- The amount of torque (T) generated at the periphery of the cutting edges of the drilling bit when it is driven at a power of (P) horsepower at some rotational speed (N).
- The relation between these quantities is given:
T = 5252*P / N
T = 5252*5.85 / 1900
T = 16.171 Nm
- The force (F) applied at the periphery of the drill bit cutting edge at a distance of radius from the center of drill bit can be determined from the definition of Torque (T) being a cross product of the Force (F) and a moment arm (r):
T = F*r
Where, r = d / 2
F = 2T / d
F = 2*16.171 / 0.98
F = 33 N
Answer: The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.
