A block with a mass of 473 g is sitting at the edge of a very smooth tabletop. T

Question

A block with a mass of 473 g is sitting at the edge of a very smooth tabletop. The height of the table is 1.15 m. A bullet with a mass of 3.70 g is then shot into the block and remains stuck in the block as it falls to the ground. After the collision, the block lands 2.35m away from the edge of the tables how much kinetic energy was lost to heat and friction during the collision between the bullet and the block? A block with a mass of 473 g is sitting at the edge of a very smooth tabletop. The height of the table is 1.15 m. A bullet with a mass of 3.70 g is then shot into the block and remains stuck in the block as it falls to the ground. After the collision, the block lands 2.35m away from the edge of the tables how much kinetic energy was lost to heat and friction during the collision between the bullet and the block?

Explanation / Answer

Heat loss = Total PE - Final KE

1.15= 0.5* g* t^2

t= 0.48 second

Range = ux* t

2.35 = ux* 0.48

ux = 4.89

KE = 1/2 mv^2 = 1/2 *0.4767* 4.89^2 = 5.71

PE = mgh = 0.476* 9.8* 1.15 = 5.36 J

Heat loss = 5.76-5.36 = 0.40 J

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