A contestant in a winter games event pushes a 54.0 kg block of ice across a frozen lake with a rope over his shoulder as shown in Figure 4.29(b). The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03.
(a) Calculate the minimum force F he must exert to get the block moving.
N
(b) What is its acceleration once it starts to move, if that force is maintained?
m/s 2
+33661 (a) Net downward vertical force on block = mass*g - F*sin(θ) or 54*9.8 - F*sin(25°) N
Resistive force = μ*(mass*g - F*sin(θ)) = 0.1*(54*9.8 - F*sin(25°) ) N
Hence, to get the block moving we must have F*cos(θ) = μ*(mass*g - F*sin(θ))
F = μ*mass*g/(cos(θ) + μ*sin(θ))
$${\mathtt{F}} = {\frac{{\mathtt{0.1}}{\mathtt{\,\times\,}}{\mathtt{54}}{\mathtt{\,\times\,}}{\mathtt{9.8}}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{25}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.1}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{25}}^\circ\right)}\right)}} \Rightarrow {\mathtt{F}} = {\mathtt{55.789\: \!263\: \!395\: \!076\: \!610\: \!5}}$$
or F ≈ 55.8 N
(b) Net force horizontal force on the block once it is moving = F*cos(θ) - μk*(mass*g - F*sin(θ))
where μk is kinetic coefficient of friction. This force equals mass*a, where a is acceleration, so:
mass*a = F*cos(θ) - μk*(mass*g - F*sin(θ))
a = F*(cos(θ) + μk*sin(θ))/mass - μk*g
$${\mathtt{a}} = {\frac{{\mathtt{55.789\: \!263\: \!395\: \!1}}{\mathtt{\,\times\,}}\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{25}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.03}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{25}}^\circ\right)}\right)}{{\mathtt{54}}}}{\mathtt{\,-\,}}{\mathtt{0.03}}{\mathtt{\,\times\,}}{\mathtt{9.8}} \Rightarrow {\mathtt{a}} = {\mathtt{0.655\: \!436\: \!494\: \!326\: \!533\: \!9}}$$
or a ≈ 0.655 m/s2
.
+33661 (a) Net downward vertical force on block = mass*g - F*sin(θ) or 54*9.8 - F*sin(25°) N
Resistive force = μ*(mass*g - F*sin(θ)) = 0.1*(54*9.8 - F*sin(25°) ) N
Hence, to get the block moving we must have F*cos(θ) = μ*(mass*g - F*sin(θ))
F = μ*mass*g/(cos(θ) + μ*sin(θ))
$${\mathtt{F}} = {\frac{{\mathtt{0.1}}{\mathtt{\,\times\,}}{\mathtt{54}}{\mathtt{\,\times\,}}{\mathtt{9.8}}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{25}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.1}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{25}}^\circ\right)}\right)}} \Rightarrow {\mathtt{F}} = {\mathtt{55.789\: \!263\: \!395\: \!076\: \!610\: \!5}}$$
or F ≈ 55.8 N
(b) Net force horizontal force on the block once it is moving = F*cos(θ) - μk*(mass*g - F*sin(θ))
where μk is kinetic coefficient of friction. This force equals mass*a, where a is acceleration, so:
mass*a = F*cos(θ) - μk*(mass*g - F*sin(θ))
a = F*(cos(θ) + μk*sin(θ))/mass - μk*g
$${\mathtt{a}} = {\frac{{\mathtt{55.789\: \!263\: \!395\: \!1}}{\mathtt{\,\times\,}}\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{25}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.03}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{25}}^\circ\right)}\right)}{{\mathtt{54}}}}{\mathtt{\,-\,}}{\mathtt{0.03}}{\mathtt{\,\times\,}}{\mathtt{9.8}} \Rightarrow {\mathtt{a}} = {\mathtt{0.655\: \!436\: \!494\: \!326\: \!533\: \!9}}$$
or a ≈ 0.655 m/s2
.
