Answer:
624.5 N
Explanation:
m = mass of the truck = 2150 kg
v₀ = initial speed of the truck = 55 km/h = 15.3 m/s
v = final speed of the truck = 33 km/h = 9.2 m/s
t = time interval = 21 s
F = magnitude of the average net force
Using Impulse-Change in momentum equation
F t = m (v - v₀ )
F (21) = (2150) (9.2 - 15.3)
F = - 624.5 N
hence the magnitude of net force is 624.5 N
The magnitude of the average net force acting on the truck during the 21.0 s interval is about 626 N
[tex]\texttt{ }[/tex]
Acceleration is rate of change of velocity.
[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]
[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]
a = acceleration ( m/s² )
v = final velocity ( m/s )
u = initial velocity ( m/s )
t = time taken ( s )
d = distance ( m )
Let us now tackle the problem!
[tex]\texttt{ }[/tex]
Given:
mass of truck = m = 2150 kg
initial speed of truck = u = 55.0 km/h = 15⁵/₁₈ m/s
final speed of truck = v = 33.0 km/h = 9¹/₆ m/s
time taken = t = 21.0 s
Asked:
average net force = ∑F = ?
Solution:
[tex]\Sigma F = m a[/tex]
[tex]\Sigma F = m ( v - u ) \div t[/tex]
[tex]\Sigma F = 2150 ( 9\frac{1}{6} - 15 \frac{5}{18} ) \div 21.0[/tex]
[tex]\Sigma F = 2150 ( -6\frac{1}{9} ) \div 21.0[/tex]
[tex]\Sigma F = -626 \texttt{ N}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Physics
Chapter: Kinematics
