A mass of m= 1.30 kg slides from an initial speed of up 1.82 m/s down...
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A mass of m= 1.30 kg slides from an initial speed of up 1.82 m/s down the inclined plane shown in Figure 1. The height of the plane is h=2.14 m and it is inclined at an angle of 37.4 to the horizontal. The coefficient of static friction between the block and the plane is the plane is μ = 0.257. 0.284. The coefficient of kinetic friction between the block and 0 Part 1) As the block slides to the bottom of the plane, what is the change in its potential energy? AU Figure 1: Diagrammatic representation of a block of mass m resting at the top of an inclined plane, which sloped upwards from left to right at an angle of 0 degrees to the horizontal. The plane has a height of h metres. Part 2) How much work is done by friction as the block slides down the plane? W = Part 3) What is the speed of the block at the bottom of the plane? 3 h A mass of m= 1.30 kg slides from an initial speed of up 1.82 m/s down the inclined plane shown in Figure 1. The height of the plane is h=2.14 m and it is inclined at an angle of 37.4 to the horizontal. The coefficient of static friction between the block and the plane is the plane is μ = 0.257. 0.284. The coefficient of kinetic friction between the block and 0 Part 1) As the block slides to the bottom of the plane, what is the change in its potential energy? AU Figure 1: Diagrammatic representation of a block of mass m resting at the top of an inclined plane, which sloped upwards from left to right at an angle of 0 degrees to the horizontal. The plane has a height of h metres. Part 2) How much work is done by friction as the block slides down the plane? W = Part 3) What is the speed of the block at the bottom of the plane? 3 h
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Solo Given 13 N h 214m m 130 kar v 182 ms 0 374 45 0284 lx M... View the full answer
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