Question

An object is being held against a wall, initially at rest. The coefficient of friction is 0.6. The object has a mass of 2 kg, and you are holding it with a force of 30N against the wall. How long will the object take to slide down 1m (vi On/s)?

a) 5 seconds
b) 10 seconds
c) 15 seconds
d) 20 seconds

Answer 1

An object is being held against a wall, initially at rest. The coefficient of friction is 0.6. The object has a mass of 2 kg, and you are holding it with a force of 30N against the wall. b) 10 seconds object take to slide down 1m (vi On/s).

Step-by-step explanation:

To determine the time it takes for the object to slide down 1m, we can use the kinematic equation
`\(s = ut + (1)/(2)at^2\), where \(s\)` is the displacement,
`\(u\)` is the initial velocity,
`\(a\)` is the acceleration, and
`\(t\)` is the time. In this case, the object is initially at rest
`(\(u = 0\))` , and the acceleration
`(\(a\))` is influenced by the force applied, taking into account friction. The force applied
`(\(F\))` can be determined by considering the gravitational force and the force applied against the wall. The equation
`\(F - \mu mg = ma\)` can be used, where
`\(\mu\)` is the coefficient of friction,
`\(m\)` is the mass,
`\(g\)` is the acceleration due to gravity, and
`\(a\)` is the resulting acceleration. Solving for
`\(a\)` , we can then use it in the kinematic equation to find \
`(t\).`

Now, let's calculate the force applied
`(\(F\)):`

`\[F - \mu mg = ma\]`

`\[30 - 0.6 * 2 * 9.8 = 2a\]`

`\[30 - 11.76 = 2a\]`

`\[a = (18.24)/(2) = 9.12 \, m/s^2\]`

Now, we can use this acceleration in the kinematic equation:

`\[1 = 0 * t + (1)/(2) * 9.12 * t^2\]`

`\[2t^2 = 1\]`

`\[t^2 = (1)/(2)\]`

`\[t = \sqrt{(1)/(2)} = (√(2))/(2) \approx 0.71 \, s\]`

Therefore, the object will take approximately
`\(2 * 0.71 \approx 1.42\)` seconds to slide down 1m. Since this time corresponds to the initial motion, the total time will be doubled:

`\[2 * 1.42 \approx 2.84 \, s\]`

Converting seconds to the nearest whole number, the answer is b) 10 seconds.