At a lumberjack competition an axe with a mass of 1.25 kg is thrown at a target
ID: 2003919 • Letter: A
Question
At a lumberjack competition an axe with a mass of 1.25 kg is thrown at a target with a speed of 41 m/s. When it hits the target, it penetrates to a depth of 0.055 m.(a) What was the average force exerted by the target on the axe?
1 N
(b) If the mass of the axe is halved, and the force exerted by the target on the axe remains the same, by what multiplicative factor does the penetration depth change?
It doubles. It increases by a factor of the square root of two. It does not change. It decreases by a factor of the square root of two. It halves.
Explanation / Answer
Given mass of axe m = 1.25 kg initial velocity of axe is u = 0 m/s axe thrown at a target with a speed of 41 m/s distance travelled by the axe after it hits the target is 0.055m From kinematic equations v 2 - u 2 = 2 as acceleration of axe is a = v 2 / 2s = ( 41 ) 2 / 2*0.055 = 15281.81 m /s 2 force exerted by the target on the axe F = ma = 1.25 * 15281.81 = 19102.272 N b) If the mass of the axe is halved, and the force exerted by the target on the axe remains the same, acceleration is a = F / m = 2 F / m = 2 (15281.81 m /s 2 ) = 30563.62 change in penetration depth s = v 2 / 2a = (41)2 / 2 ( 30563.62 ) = 0 .0275 m b) If the mass of the axe is halved, and the force exerted by the target on the axe remains the same, acceleration is a = F / m = 2 F / m = 2 (15281.81 m /s 2 ) = 30563.62 change in penetration depth s = v 2 / 2a = (41)2 / 2 ( 30563.62 ) = 0 .0275 m = (41)2 / 2 ( 30563.62 ) = 0 .0275 mRelated Questions
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