A sphere of mass 0.5 kg rests at one corner of a cube. The cube is moved with a velocity v = 8 t i ^ − 2 t 2 j ^ where t is time in seconds. Find the magnitude of force by sphere on the cube at t =

Question

A sphere of mass 0.5 kg rests at one corner of a cube. The cube is moved with a velocity $v = 8 t \hat{i} - 2 t^{2} \hat{j}$ where t is time in seconds. Find the magnitude of force by sphere on the cube at $t = 1 s$ is [figure shows vertical plane of the cube, $g = 10 m / s^{2}$ ]

$= 1 s g = 10 m / s^{2}]$

$v = 8 t \hat{i} - 2 t^{2} \hat{j}$

TestHub · Accepted Answer

a = d t d v ​ = 8 i ^ − 4 t j ^ ​ ; at t = 1 s: 8 i ^ − 4 j ^ ​ m/s 2 The cube pushes the sphere horizontally: N x ​ = m a x ​ = 0.5 × 8 = 4 N Vertically, with the cube accelerating down at 4: N y ​ = m ( g − 4 ) = 0.5 × 6 = 3 N ∣ F ∣ = 4 2 + 3 2 ​ = 5 N

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