Menu div.question,div.steps {font-size: 16px; padding: 10px;line-height: 20px; margin: 20px;} div.question{border: 1px solid #ccc;} div#solution .button{font-size: 20px; margin: 20px; width: 200px; padding: 10px; text-align: center;} div.hint,div.answer,div.steps{display: none;} div.pre_next{font-size: 30px; width:90%; margin:auto;} div.pre_next a.nav_pre{float:left;} div.pre_next a.nav_next{float:right;} div.title div.notes{ font-family:sans-serif, Arial; font-size: 18px; } div.speak{display:none;} i.speak{float: right;margin-top:-10px;cursor:pointer;} 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`