Want this question answered?
F=ma. Greater mass leads to smaller acceleration given the same force.
force = mass x acceleration so if force doubles acceleration doubles to 8 m/s2
Use Newton's Second Law, F=ma. Solving for a: a = F/m (acceleration = force / mass). If the force is in Newton, and the mass in kilograms, acceleration will be in meters/second2.
.5m\s2
The cart's acceleration will decrease as its mass increases. This is why you must exert progressively more force on a shopping cart to move it along as items are added to it. If you were to continue to add items to the cart but not change how hard you push it, the cart would eventually become "impossible" to push.
F=ma. Greater mass leads to smaller acceleration given the same force.
force = mass x acceleration so if force doubles acceleration doubles to 8 m/s2
The cart's acceleration will be directly proportional to the net force applied to it. If the force remains constant, the acceleration will also remain constant, assuming no other external factors are affecting the cart's motion.
The basic equation is: force equals mass times acceleration.
Use Newton's Second Law, F=ma. Solving for a: a = F/m (acceleration = force / mass). If the force is in Newton, and the mass in kilograms, acceleration will be in meters/second2.
F=mass * acceleration 60kg m/s^2=10kg * acceleration 6m/s^2 = acceleration
.5m\s2
The cart's acceleration will decrease as its mass increases. This is why you must exert progressively more force on a shopping cart to move it along as items are added to it. If you were to continue to add items to the cart but not change how hard you push it, the cart would eventually become "impossible" to push.
(Force on an object) = (the object's mass) times (its acceleration)
According to Newton's Second Law of Motion, the greater the force, the greater the acceleration. So if you were to begin pushing a shopping cart harder, you go faster and there is more acceleration. If you were to push the cart softer there would be less acceleration.
That depends on the force applied.
Doubled.