7. A `10`-`kg` block slides down form `3.0`-`m`-height and `10`-`m`-long inclined plane as shown. It reaches the bottom with a speed of `4.0\ m//s`, Determine the friction between the block and the inclined surface.
7. A 10-kg block slides down form 3.0-m-height and 10-m-long inclined plane as shown. It reaches the bottom with a speed of 4.0\ m//s, Determine the friction between the block and the inclined surface.
Hint
`KE_i + PE_i = KE_r + PE_r + W_(fr)`
Answer
`28\ N`
Show Steps
Given: `m = 10\ kg` `h = 3.0\ m` `d = 10\ m` `v_f = 4.0\ m//s` Equations: `W = Fd` `KE_i + PE_i = KE_f + PE_f + W_(f r)` Solution: `KE_i + PE_i = KE_i + PE_i + W_(f r)` `0 + mgh = 1/2 mv_f^2 + 0 + W_(f r)` `W_(f r) = mgh -1/2 mv_f^2` `W_(f r) = m(gh -1/2v_f^2)` `W_(f r) = (10\ kg)((10\ m//s^2)(3.0\ m) -1/2(4.0\ m//s)^2)` `W_(f r) = 220 \ J` `W_(fr) = f_kd
`f_k = W_(fr)/d = (220\ J) / (10\ m) = 22\ N`