A textbook with a mass of 2.18 kg is pulled across a table with a horizontal force of 5.5 N [E]. If the table has a coefficient of friction of 0.15, what is the acceleration of the textbook?
1 month ago
acceleration = 1.1 m/s2
Fnet = Fapplied - Ffriction
Fnet = mass*acceleration; Fapplied = 5.5 N; Ff = μ*FNormal --> FNormal = Fgravity
Fnet = 5.5N - (0.15*9.81m/s2 *2.18kg) --> 5.5N - 3.21N
Fnet (m*a) = 2.29 N
accel = 2.29 N/mass --> 2.29 (kg*m/s2) /2.18kg -->1.05 m/s2 --> rounds to 1.1 m/s2
1 month ago
Answered By Tara K
acceleration = 1.1 m/s2
Fnet = Fapplied - Ffriction
Fnet = mass*acceleration; Fapplied = 5.5 N; Ff = μ*FNormal --> FNormal = Fgravity
Fnet = 5.5N - (0.15*9.81m/s2 *2.18kg) --> 5.5N - 3.21N
Fnet (m*a) = 2.29 N
accel = 2.29 N/mass --> 2.29 (kg*m/s2) /2.18kg -->1.05 m/s2 --> rounds to 1.1 m/s2