Answer :
Answer:
a. 35N
b. 1.37N
Explanation:
(a) The force on the block can be obtained with the second Newton's Law
[tex]F=ma\\F=(14kg)(2.5\frac{m}{s^2})=35N[/tex]
(b)
we can assume that the pull has the same acceleration. Hence we have
[tex]F_{pull}=(0.55kg)(2.5\frac{m}{s^2})=1.37N[/tex]
Hope this helps!!
Answer:
a) 35 N force pulls on the Ice
b) 36.375 N force pulls on the rope
Explanation:
Mass of the rope, [tex]m_{R} = 550g = 0.55 kg[/tex]
Length of the rope, [tex]l_{R} = 1.50 m[/tex]
Mass of the Ice, [tex]m_{I} = 14.0 kg[/tex]
Acceleration of the ice block, [tex]a_{I} = 2.50 m/s^{2}[/tex]
a) The amount of force that pulls forward on the Ice, [tex]F_{I}[/tex]
[tex]F_{I} = m_{I} a_{I} \\F_{I} = 14 * 2.5\\F_{I} = 35 N[/tex]
b) The amount of force that pulls forward on the rope, [tex]F_{R}[/tex]
Since the rope is pulling the ice, both the force on the ice and the force on the rope alone combine to pull the rope.
Moreover, both the ice and the rope accelerate equally, [tex]a_{R} = 2.5 m/s^{2}[/tex]
[tex]F_{R} = 35 + m_{R} a_{R} \\F_{R} = 35 + (0.55*2.5)\\F_{R} = 36.375 N[/tex]