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?
5 months 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
To find acceleration, we need to determine the net force acting on the textbook. There are two horizontal forces: the pulling force and friction.
Step 1: Calculate friction force
Step 2: Calculate net force
Step 3: Calculate acceleration (F = ma)
5 months 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
1 month ago
Answered By Dhruvin P
To find acceleration, we need to determine the net force acting on the textbook. There are two horizontal forces: the pulling force and friction.
Step 1: Calculate friction force
Normal force = mass × gravity = 2.18 kg × 9.81 m/s² = 21.39 N Friction force = μ × Normal force = 0.15 × 21.39 = 3.21 NStep 2: Calculate net force
Net force = Applied force - Friction force = 5.5 N - 3.21 N = 2.29 NStep 3: Calculate acceleration (F = ma)
a = F/m = 2.29 N ÷ 2.18 kg = 1.05 m/s²