A 4.20–kg block is set into motion up an inclined plane with an initial speed of vi = 7.40 m/s (see figure below). The block comes to rest after traveling d = 3.00 m along the plane, which is inclined at an angle of 𝜃 = 30.0° to the horizontal.
Answer
By energy conservation
So
"mgh+\\frac{1}{2}mv^2=\\mu mgcos\\theta d"
"9.8(3.00sin30\u00b0)+\\frac{1}{2}(7.40)^2=\\mu (9.8)(cos30\u00b0)(3)"
So
"\\mu=1.65"
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