The velocity of a 1.5 kg blocksliding down a frictionless inclined plane is foun

Question

The velocity of a 1.5 kg blocksliding down a frictionless inclined plane is found to be 1.51 m/s.1.40 s later, it has a velocity of 5.85 m/s.
What is the angle of the plane with respect to the horizontal (indegrees)? Can I get a numerical answer? The velocity of a 1.5 kg blocksliding down a frictionless inclined plane is found to be 1.51 m/s.1.40 s later, it has a velocity of 5.85 m/s.
What is the angle of the plane with respect to the horizontal (indegrees)? Can I get a numerical answer? Can I get a numerical answer?

Explanation / Answer

Initial speed (u) of the block = 1.51 m/s Final speed (v) of the block = 5.85 m/s Time (t) taken = 1.40s Now acceleration (a) of the block is                  a = (v - u)/t                     = (5.85m/s - 1.51m/s) / (1.40s)                     = 3.1 m/s2When the block is sliding down theinclined plane the acceleration (a) of the block is                  a = gsin             3.1 m/s2 = (9.8m/s2)sin                  = 18.4o

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