**Physics Problem Set****7. An elevator of mass 890 kg has an upward force of 1.10 x 10⁴ N applied to it. Determine the following:**a) The net force on the elevator.b) The acceleration of the elevator.*(ANS: a) 2.23 x 10³ N b) 2.56 m/s²)***8. In a competition, a small robot accelerates forward at 0.25 m/s² as it pushes a block of wood with a = 0.25 m/s² and a mass of 1.6 kg towards a basket. A friction force of 3.7 N exists between the block and the floor.**a) What is the net force acting on the block?b) What force is the robot applying to accelerate the block?*(ANS: a) 0.4 N b) 4.1 N)**Diagram Explanation:*The diagram depicts a robot pushing a block forward. The block is shown on a horizontal surface with an arrow indicating the direction of acceleration, which is 0.25 m/s².**9. A hoist is used to lift a 780 kg car upward with an acceleration of 0.110 m/s².**a) What is the net force acting on the car?b) What force was used to lift the car?*(ANS: a) 86 N b) 7.73 x 10³ N)**Diagram Explanation:*The diagram illustrates a car being lifted by a hoist. The car is positioned on a platform with an upward arrow indicating the acceleration of 0.110 m/s².

7 -- Given: Mass of elevator, m=890 kg Upward force , F=1.10×104 N To determine: (a) net force on the elevator. (b) the acceleration of the elevator. Note: According to Bartleby q&a guidelines, the experts are advised to solve only first question if multiple different questions are posted in a single post. Please post remaining questions separately.(a) Upward force of the elevator is F==1.10×104 N The downward force, weight on the elevator is Weight, W = mg Where m is mass and g is acceleration due to gravity. Putting m=890 kg and g=9.8 m/s2 W=(890 kg)(9.8 m/s2) W=8722 N Since upward force is greater than downward force (weight) the elevator will upward, the net force on the elevator is Fnet = F - W F=1.10×104 N and W=8722 N Fnet = 1.10×104 N - 8722 N Fnet =2.278×103 N The net force on the elevator is 2.278×103 N. (b) The net acceleration is Fnet = ma a = Fnet/m Where m is mass of elevator, m=890 kg Fnet =2.278×103 N a = (2.278×103 N)/(890 kg) a = 2.559 m/s2 a = 2.56 m/s2 The net acceleration of the elevator is 2.56 m/s2.

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N7. An elevator of mass 890 kg has an upward force of 1.10 x 10N applied to it. Determine the following: a) the net force on the elevator. b) the acceleration of the elevator. (ANS: a) 2.23 x 103 N b) 2.56 m/s²) 8. In a comnetition

N7. An elevator of mass 890 kg has an upward force of 1.10 x 10N applied to it. Determine the following: a) the net force on the elevator. b) the acceleration of the elevator. (ANS: a) 2.23 x 103 N b) 2.56 m/s²) 8. In a comnetition

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Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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Question

**Physics Problem Set**

**7. An elevator of mass 890 kg has an upward force of 1.10 x 10⁴ N applied to it. Determine the following:**

a) The net force on the elevator.

b) The acceleration of the elevator.

*(ANS: a) 2.23 x 10³ N b) 2.56 m/s²)*

**8. In a competition, a small robot accelerates forward at 0.25 m/s² as it pushes a block of wood with a = 0.25 m/s² and a mass of 1.6 kg towards a basket. A friction force of 3.7 N exists between the block and the floor.**

a) What is the net force acting on the block?

b) What force is the robot applying to accelerate the block?

*(ANS: a) 0.4 N b) 4.1 N)*

*Diagram Explanation:*
The diagram depicts a robot pushing a block forward. The block is shown on a horizontal surface with an arrow indicating the direction of acceleration, which is 0.25 m/s².

**9. A hoist is used to lift a 780 kg car upward with an acceleration of 0.110 m/s².**

a) What is the net force acting on the car?

b) What force was used to lift the car?

*(ANS: a) 86 N b) 7.73 x 10³ N)*

*Diagram Explanation:*
The diagram illustrates a car being lifted by a hoist. The car is positioned on a platform with an upward arrow indicating the acceleration of 0.110 m/s².

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